Solution:
Let $w,x,y,z$ denote the number of residents who own 1,2,3 and 4 of these items, respectively. We know $w+x+y+z=900$, since there are 900 residents in total. This simplifies to
$w+z=229$, since we know $x=437$ and $y=234$.
Now, we set an equation of the total number of items. We know there are 195 rings, 367 clubs, 562 spades, and 900 candy hearts. Adding these up, there are 2024 (wow! the year!) items in total. Thus, $w+2x+3y+4z=2024$ since we are adding the number of items each group of people contributes, and this must be equal to the total number of items.
Plugging in x and y once more, we get $w+4z=448$. Solving $w+z=229$ and $w+4z=448$, we get $z=\boxed{073}$
-Westwoodmonster
Let $a,b,c$ denote the number of residents that own only a diamond ring and a bag of candy hearts, the number of residents that own only a golf club and a bag of candy hearts, and the number of residents that own only a garden spade and a bag of candy hearts, respectively. Let $x,y,z$ denote the number of residents that own only a diamond ring, a golf club, and a bag of candy hearts; the number of residents that own only a diamond ring, a garden spade, and a bag of candy hearts; and the number of residents that own only a golf club, a garden spade, and a bag of candy hearts. Let $n$ denote the number of people that own all $4$ items.
$a+x+y+n=195$ (the number of people that got diamond rings), $b+x+z+n=367$ (the number of people that got golf clubs), $c+y+z+n=562$ (the number of people that got garden spades). We also know $a+b+c=437$ (the number of people that own two objects), and $x+y+z=234$ (the number of people that own three objects). Adding the first three equations gives \[a+b+c+2(x+y+z)+3n=1124.\]
Substituting the second two equations gives $437+2\cdot 234+3n=1124$, so $n=\boxed{073}$.
~nezha33
We know that there are 195 diamond rings, 367 golf clubs, and 562 garden spades, so we can calculate that there are $195+367+562=1124$ items, with the exclusion of candy hearts which is irrelevant to the question. There are 437 people who owns 2 items, which means 1 item since candy hearts are irrelevant, and there are 234 people who own 2 items plus a bag of candy hearts, which means that the 234 people collectively own $234*2=468$ items. We can see that there are $1124-437-468=219$ items left, and since the question is asking us for the people who own 4 items, which means 3 items due to the irrelevance of candy hearts, we simply divide 219 by 3 and get $219/3=\boxed{073}$.
~Callisto531
Let $a$ be the number of people who have exactly one of these things and let $b$ be the number of people who have exactlty four of these objects. We have $a + 437 + 234 + d = 900,$ so $a + d = 229$.

Including those who have more than one object, we have
\[195 + 367 + 562 + 900 = a + 2\cdot 437 + 3\cdot 234 + 4d.\]
This is because we count those who own exactly $2$ objects twice, those who own $3$ thrice, and those who own $4$ four times. Solving gives $a + 4d = 448$.

Solving the system $a + 4d = 448, a + d = 229$ gives $3d = 219,$ so $d = \boxed{\textbf{(073)}}$.

-Benedict T (countmath1)

Answer: 073

<｜begin▁of▁sentence｜><｜User｜>Among the 900 residents of Aimeville, there are 195 who own a diamond ring, 367 who own a set of golf clubs, and 562 who own a garden spade. In addition, each of the 900 residents owns a bag of candy hearts. There are 437 residents who own exactly two of these things, and 234 residents who own exactly three of these things. Find the number of residents of Aimeville who own all four of these things.<｜Assistant｜><think>

Okay, so I need to solve this problem about the residents of Aimeville and the things they own. Let me read it again and make sure I understand all the numbers given.

There are 900 residents total. Each of them owns a bag of candy hearts. So, candy hearts are universal here—everyone has those. Then, there are three other items: diamond rings, golf clubs, and garden spades. The numbers given are 195 for diamond rings, 367 for golf clubs, and 562 for garden spades.

Additionally, it says that 437 residents own exactly two of these things (excluding the candy hearts, I assume, since everyone has those). And 234 residents own exactly three of these things. The question is asking for the number of residents who own all four things.

Hmm, okay. Since everyone owns the candy hearts, the "all four things" would mean they own the candy hearts plus the other three items. So, the problem is essentially about overlapping ownership of the three items (diamond ring, golf clubs, garden spade) on top of the universal candy hearts.

This seems like a classic inclusion-exclusion problem. Let me recall the formula for four sets, but since candy hearts are owned by everyone, maybe we can simplify it? Wait, actually, since candy hearts are universal, maybe we can treat the problem as three sets (the other three items) over the universe of 900 people, and the "exactly two" and "exactly three" refer to the other three items.

Let me clarify the problem again. The four things are: diamond ring, golf clubs, garden spade, and candy hearts. But since everyone has candy hearts, the ownership of the four things would just be the intersection of the other three with the candy hearts, but since everyone has candy hearts, it's equivalent to just the intersection of the other three.

Therefore, the problem reduces to three sets (D, G, S) over 900 elements, where:

- |D| = 195

- |G| = 367

- |S| = 562

Additionally, the number of people who own exactly two of D, G, S is 437, and exactly three of D, G, S is 234. We need to find the number of people who own all three, which would be the intersection of D, G, and S, right? Since owning all four things (including candy hearts) is the same as owning all three of the other items because candy hearts are universal.

Wait, but the problem says "the number of residents who own all four of these things." Since all four includes the candy hearts, which everyone has, so yes, owning all four is equivalent to owning all three of the other items. So the answer is the number of people who own all three of D, G, and S.

So, let me rephrase the problem in terms of three sets. Let me denote:

Let D be the set of diamond ring owners,

G be the set of golf club owners,

S be the set of garden spade owners.

We need to find |D ∩ G ∩ S|.

Given:

Total population, N = 900.

Each person is in at least the universal set, but since candy hearts are universal, the other sets are subsets of the 900.

Given:

|D| = 195,

|G| = 367,

|S| = 562,

Number of people with exactly two of D, G, S is 437,

Number with exactly three of D, G, S is 234.

We need to find |D ∩ G ∩ S|.

Okay, so in inclusion-exclusion terms, the total number in at least one of the sets D, G, S is equal to the sum of the individual sets minus the sum of all pairwise intersections plus the triple intersection. But here, since everyone has candy hearts, but the other items are optional. Wait, actually, the problem says "each of the 900 residents owns a bag of candy hearts," so the other items are in addition? Wait, hold on. Wait, actually, the problem says "among the 900 residents... there are 195 who own a diamond ring, 367 who own a set of golf clubs, and 562 who own a garden spade. In addition, each of the 900 residents owns a bag of candy hearts."

So, the total ownerships are: everyone has candy hearts, and some have the other items. So the other items (diamond, golf, garden) are in addition to the candy hearts. Therefore, the four things are separate, but candy hearts are universal. So when they say "own exactly two of these things," they mean exactly two of the four, but since everyone has candy hearts, then "exactly two" would mean one of the other three plus the candy hearts. Wait, but that complicates things. Wait, maybe I need to parse the problem again carefully.

Wait, the problem states: "there are 195 who own a diamond ring, 367 who own a set of golf clubs, and 562 who own a garden spade. In addition, each of the 900 residents owns a bag of candy hearts."

So, the four things are diamond ring, golf clubs, garden spade, and candy hearts. Each resident owns at least the candy hearts. The counts for the other three are given. So, the total number of residents is 900, each has at least the candy hearts, and some have the other items.

Now, the problem says "there are 437 residents who own exactly two of these things," meaning exactly two of the four (diamond, golf, garden, candy hearts). But since everyone owns candy hearts, owning exactly two would mean owning candy hearts plus exactly one of the other three. Similarly, "exactly three of these things" would mean candy hearts plus exactly two of the other three. And "exactly four" would mean candy hearts plus all three of the other items.

Therefore, the counts given for exactly two and exactly three refer to the total number of items owned, including candy hearts. Therefore, to model this, perhaps it's better to think in terms of the three non-universal items (D, G, S) and consider how many people have 0, 1, 2, or 3 of those, since the candy hearts are fixed.

Let me formalize this:

Let’s denote:

- Let x be the number of people who own none of D, G, S (only candy hearts).

- Let a be the number of people who own exactly one of D, G, S.

- Let b be the number of people who own exactly two of D, G, S.

- Let c be the number of people who own all three of D, G, S.

Then, the total population is x + a + b + c = 900.

But the problem states that the number of residents who own exactly two of these things (i.e., exactly two of D, G, S plus candy hearts, so total of three items including candy hearts) is 437. Wait, hold on. Wait, the problem says "exactly two of these things"—the "these things" refers to all four items (diamond, golf, garden, candy hearts). Since each resident owns the candy hearts, so "exactly two of these things" would mean exactly one of the other three items plus the candy hearts. Therefore, the count of people with exactly two things is the number of people with exactly one of D, G, S. Similarly, "exactly three of these things" would be exactly two of D, G, S plus candy hearts, so that's the number of people with exactly two of D, G, S. And "exactly four" would be those with all three of D, G, S plus candy hearts, so that's c.

Wait, this is a bit confusing. Let me parse the problem again:

The problem says: "there are 437 residents who own exactly two of these things, and 234 residents who own exactly three of these things."

The phrase "these things" must refer to the four items mentioned: diamond ring, golf clubs, garden spade, and candy hearts. Since each resident owns the candy hearts, then:

- Owning exactly two of these things means owning exactly one of the other three (since candy hearts are already one).

- Owning exactly three of these things means owning exactly two of the other three.

- Owning all four would mean owning all three of the other items plus the candy hearts.

Therefore, the given numbers are:

- Exactly two things: 437 → this corresponds to exactly one of D, G, S.

- Exactly three things: 234 → this corresponds to exactly two of D, G, S.

- Exactly four things: ? → this is exactly three of D, G, S, which is c.

- Exactly one thing: these would be the people who only have candy hearts, so x (since they have zero of D, G, S). Wait, but "exactly one thing" would be x, since they have only the candy hearts.

Therefore, the total population is:

x (exactly one thing: only candy hearts) +

a (exactly two things: one of D, G, S plus candy hearts) +

b (exactly three things: two of D, G, S plus candy hearts) +

c (exactly four things: all three of D, G, S plus candy hearts) =

x + a + b + c = 900.

But according to the problem, the number of people with exactly two things (a) is 437, and exactly three things (b) is 234. So:

a = 437,

b = 234,

and we need to find c (which is the number of people with exactly four things, i.e., all three of D, G, S).

Additionally, we can also relate this to the counts of the individual items:

The total number of diamond ring owners is 195. Each diamond ring owner is in either the group that has only D, or D and one other, or D and two others, or all three. Similarly for G and S.

In terms of the variables:

The total number of D owners is:

(number with only D) + (number with D and one other) + (number with D and two others) + (number with all three).

But in terms of our variables:

Let me think. Let me denote:

Let’s define:

- Let’s denote the number of people with exactly one item (other than candy hearts) as a, but broken down:

a = (number with only D) + (number with only G) + (number with only S).

Similarly, the number of people with exactly two items (other than candy hearts) would be b, but broken down into pairs:

b = (number with D and G only) + (number with D and S only) + (number with G and S only).

And c is the number with all three, D, G, and S.

Therefore, the total number of D owners is:

(only D) + (D and G only) + (D and S only) + (all three).

Similarly for G and S.

Therefore, if we let:

Let me define variables:

Let’s denote:

- Only D: d,

- Only G: g,

- Only S: s,

- D and G only: dg,

- D and S only: ds,

- G and S only: gs,

- All three: dgs.

Then,

a = d + g + s = 437,

b = dg + ds + gs = 234,

c = dgs = ? (this is what we need to find).

Total people: (d + g + s) + (dg + ds + gs) + dgs + x = 900,

where x is the number of people with only candy hearts (none of D, G, S).

So, substituting the known values:

437 + 234 + c + x = 900,

so x = 900 - (437 + 234 + c) = 900 - 671 - c = 229 - c.

Now, the total number of D owners is:

d + dg + ds + dgs = 195,

Similarly,

Total G owners: g + dg + gs + dgs = 367,

Total S owners: s + ds + gs + dgs = 562.

So, we have three equations:

1. d + dg + ds + dgs = 195,

2. g + dg + gs + dgs = 367,

3. s + ds + gs + dgs = 562.

Also, we have:

a = d + g + s = 437,

and

b = dg + ds + gs = 234,

and c = dgs.

We can try to express the totals in terms of a, b, c, and x.

Alternatively, let me see if we can write equations in terms of the totals.

Let me denote the total D, G, S owners as:

Total D = d + dg + ds + dgs = 195,

Total G = g + dg + gs + dgs = 367,

Total S = s + ds + gs + dgs = 562.

If we add all three totals together:

Total D + Total G + Total S = (d + g + s) + 2*(dg + ds + gs) + 3*dgs.

Because:

Each of d, g, s appears once in their own total,

Each pair (dg, ds, gs) appears in two totals,

And dgs appears in all three totals.

Therefore,

195 + 367 + 562 = (d + g + s) + 2*(dg + ds + gs) + 3*dgs.

Plugging in the known values:

195 + 367 + 562 = 437 + 2*234 + 3*c.

Let me compute left side:

195 + 367 = 562; 562 + 562 = 1124. Wait, 195 + 367 is actually 562? Wait, 195 + 367: 195 + 300 is 495, plus 67 is 562. Then 562 + 562 is 1124. So left side is 1124.

Right side: 437 + 2*234 + 3c.

Compute 2*234 = 468,

So 437 + 468 = 905,

Thus, 905 + 3c = 1124,

Therefore, 3c = 1124 - 905 = 219,

So c = 219 / 3 = 73.

Wait, so that would give c = 73. Hmm, so is that the answer?

Wait, let me check my steps again to make sure.

First, the formula:

Total D + Total G + Total S = (sum of only one) + 2*(sum of exactly two) + 3*(all three).

Yes, because:

Each person who is in exactly one set contributes to one total,

Each person in exactly two sets contributes to two totals,

Each person in all three contributes to all three totals.

Therefore, the formula is correct.

Plugging in numbers:

Total D + G + S = 195 + 367 + 562 = 1124,

Sum of only one (a) is 437,

Sum of exactly two (b) is 234,

Thus,

1124 = 437 + 2*234 + 3c,

Calculates to 437 + 468 = 905,

1124 - 905 = 219,

219 = 3c ⇒ c = 73.

Therefore, the number of people who own all three of D, G, S is 73, which would mean they own all four items (including candy hearts).

But let me verify this with another approach to be sure.

Alternatively, let's use variables:

Let me denote:

We have:

d + g + s = 437,

dg + ds + gs = 234,

dgs = c.

Total D = d + dg + ds + dgs = 195,

Similarly for G and S.

Let me express each total in terms of variables.

Let me write equations for each total:

For D:

d + (dg + ds) + dgs = 195,

But dg + ds is equal to (dg + ds + gs) - gs = 234 - gs,

Wait, maybe another way. Alternatively, note that:

d = (only D) = total D - (D and G only) - (D and S only) - (all three),

Similarly,

d = 195 - (dg + ds + dgs),

Wait, no, actually:

Wait, D total is:

only D + (D and G only) + (D and S only) + all three,

so,

d + dg + ds + dgs = 195,

Similarly,

g + dg + gs + dgs = 367,

s + ds + gs + dgs = 562.

If we add all three equations:

(d + g + s) + (dg + dg + ds + ds + gs + gs) + 3*dgs = 195 + 367 + 562,

Wait, actually, when adding the three equations:

Left side:

(d + g + s) + (dg + ds + dg + gs + ds + gs) + 3*dgs,

Wait, no, let me do term by term:

First equation: d + dg + ds + dgs,

Second: g + dg + gs + dgs,

Third: s + ds + gs + dgs,

Adding them together:

(d + g + s) + (dg + dg) + (ds + ds) + (gs + gs) + 3*dgs,

Which is:

(d + g + s) + 2*(dg + ds + gs) + 3*dgs,

Which is exactly the same as before. So that gives the same equation. Therefore, the calculation is correct.

Therefore, c = 73.

But let me check if this makes sense with the total population.

We had:

Total people = a + b + c + x = 437 + 234 + 73 + x = 437 + 234 is 671, plus 73 is 744, so x = 900 - 744 = 156.

So x is 156, which is the number of people with only candy hearts.

Let me check if the total numbers for D, G, S add up.

Let me suppose c = 73.

Then, for D:

Total D = d + dg + ds + 73 = 195,

So d + dg + ds = 195 - 73 = 122.

Similarly, for G:

g + dg + gs = 367 - 73 = 294,

And for S:

s + ds + gs = 562 - 73 = 489.

Also, we know that:

d + g + s = 437,

and dg + ds + gs = 234.

Let me see if these can be consistent.

Let me denote:

Let me call the sum of d + g + s = 437,

and the sum of (d terms in D) + (g terms in G) + (s terms in S) ?

Wait, perhaps another way. Let me think of the variables:

Let me denote:

Let me consider the total of (d + dg + ds) for D is 122,

Similarly, for G: (g + dg + gs) = 294,

For S: (s + ds + gs) = 489.

If we add these three:

(d + dg + ds) + (g + dg + gs) + (s + ds + gs) = 122 + 294 + 489,

Left side:

d + g + s + 2*dg + 2*ds + 2*gs,

Which is (d + g + s) + 2*(dg + ds + gs) = 437 + 2*234 = 437 + 468 = 905,

Right side: 122 + 294 = 416; 416 + 489 = 905.

So that checks out.

Therefore, the calculations are consistent.

Therefore, the number of people who own all four things is 73.

Wait, but let me just confirm once more. Since the problem says "the number of residents who own all four of these things," which is exactly c, which we found as 73.

Therefore, I think that is the answer.

But let me just think if there's another way someone might approach this and possibly make a mistake. For instance, maybe miscounting the exactly two and three.

Alternatively, using inclusion-exclusion formula for three sets:

The formula is:

|D ∪ G ∪ S| = |D| + |G| + |S| - |D ∩ G| - |D ∩ S| - |G ∩ S| + |D ∩ G ∩ S|.

But here, the union D ∪ G ∪ S would be the number of people who own at least one of D, G, S. Since everyone has candy hearts, but some might not have any of the other three. The total population is 900, so |D ∪ G ∪ S| = 900 - x, where x is the number with none (only candy hearts).

We can also express |D ∪ G ∪ S| as the sum of those with exactly one, two, three of D, G, S:

Which is a + b + c = 437 + 234 + 73 = 744. Therefore, |D ∪ G ∪ S| = 744, so x = 900 - 744 = 156, which matches earlier.

Alternatively, using inclusion-exclusion:

|D ∪ G ∪ S| = 195 + 367 + 562 - (|D ∩ G| + |D ∩ S| + |G ∩ S|) + |D ∩ G ∩ S|.

But we need to express the pairwise intersections.

Note that |D ∩ G| includes those who have D and G and possibly S,

Similarly, |D ∩ G| = dg + dgs,

Similarly, |D ∩ S| = ds + dgs,

|G ∩ S| = gs + dgs,

Therefore,

|D ∩ G| + |D ∩ S| + |G ∩ S| = (dg + ds + gs) + 3*dgs,

Wait, no:

Wait, each pairwise intersection is:

|D ∩ G| = dg (exactly D and G) + dgs (all three),

Similarly for the others,

Thus,

|D ∩ G| + |D ∩ S| + |G ∩ S| = (dg + dgs) + (ds + dgs) + (gs + dgs) = (dg + ds + gs) + 3*dgs = 234 + 3*73 = 234 + 219 = 453.

Therefore,

|D ∪ G ∪ S| = 195 + 367 + 562 - 453 + 73,

Calculates as:

195 + 367 = 562,

562 + 562 = 1124,

1124 - 453 = 671,

671 + 73 = 744,

Which matches the previous value.

Therefore, this also confirms that the value of c = 73 is correct.

Therefore, I think the answer is 73.

**Final Answer**
\boxed{73}
</think>

Given the problem of finding the number of residents in Aimeville who own all four items (diamond ring, golf clubs, garden spade, and candy hearts), we start by noting that every resident owns a bag of candy hearts. Therefore, we need to determine the number of residents who own all three of the other items (diamond ring, golf clubs, and garden spade).

We denote:
- \( |D| = 195 \) (diamond ring owners),
- \( |G| = 367 \) (golf clubs owners),
- \( |S| = 562 \) (garden spade owners),
- 437 residents own exactly two of the three items (excluding candy hearts),
- 234 residents own exactly three of the three items (excluding candy hearts).

We need to find \( |D \cap G \cap S| \), the number of residents who own all three items (and thus all four including candy hearts).

Using the principle of inclusion-exclusion for three sets, we know:
\[
|D \cup G \cup S| = |D| + |G| + |S| - |D \cap G| - |D \cap S| - |G \cap S| + |D \cap G \cap S|
\]

We also know the total number of residents is 900, and the number of residents who own exactly two items is 437, and exactly three items is 234. Let \( c = |D \cap G \cap S| \). The total number of residents owning at least one of the three items (excluding candy hearts) is:
\[
|D \cup G \cup S| = 900 - x
\]
where \( x \) is the number of residents who own only candy hearts.

The total number of residents owning exactly one, two, or three items (excluding candy hearts) is:
\[
437 + 234 + c = 671 + c
\]
Thus, \( x = 900 - (671 + c) = 229 - c \).

Using the inclusion-exclusion principle, we sum the individual sets:
\[
|D| + |G| + |S| = 195 + 367 + 562 = 1124
\]

This sum can also be expressed as:
\[
(|D| + |G| + |S|) = ( \text{sum of exactly one} ) + 2 \times ( \text{sum of exactly two} ) + 3 \times ( \text{sum of exactly three} )
\]
Substituting the known values:
\[
1124 = 437 + 2 \times 234 + 3c
\]
Solving for \( c \):
\[
1124 = 437 + 468 + 3c \\
1124 = 905 + 3c \\
3c = 219 \\
c = 73
\]

Thus, the number of residents who own all four items is \(\boxed{73}\).


Solution:
Start off by (of course) drawing a diagram! Let $I$ and $O$ be the incenter and circumcenters of triangle $ABC$, respectively. Furthermore, extend $AI$ to meet $BC$ at $L$ and the circumcircle of triangle $ABC$ at $D$.

We'll tackle the initial steps of the problem in two different manners, both leading us to the same final calculations.
Since $I$ is the incenter, $\angle BAL \cong \angle DAC$. Furthermore, $\angle ABC$ and $\angle ADC$ are both subtended by the same arc $AC$, so $\angle ABC \cong \angle ADC.$ Therefore by AA similarity, $\triangle ABL \sim \triangle ADC$.
From this we can say that \[\frac{AB}{AD} = \frac{AL}{AC} \implies AB \cdot AC = AL \cdot AD\]
Since $AD$ is a chord of the circle and $OI$ is a perpendicular from the center to that chord, $OI$ must bisect $AD$. This can be seen by drawing $OD$ and recognizing that this creates two congruent right triangles. Therefore, \[AD = 2 \cdot ID \implies AB \cdot AC = 2 \cdot AL \cdot ID\]
We have successfully represented $AB \cdot AC$ in terms of $AL$ and $ID$. Solution 1.2 will explain an alternate method to get a similar relationship, and then we'll rejoin and finish off the solution.
$\angle ALB \cong \angle DLC$ by vertical angles and $\angle LBA \cong \angle CDA$ because both are subtended by arc $AC$. Thus $\triangle ABL \sim \triangle CDL$.
Thus \[\frac{AB}{CD} = \frac{AL}{CL} \implies AB = CD \cdot \frac{AL}{CL}\]
Symmetrically, we get $\triangle ALC \sim \triangle BLD$, so
\[\frac{AC}{BD} = \frac{AL}{BL} \implies AC = BD \cdot \frac{AL}{BL}\]
Substituting, we  get \[AB \cdot AC = CD \cdot \frac{AL}{CL} \cdot BD \cdot \frac{AL}{BL}\]
Lemma 1: BD = CD = ID
Proof:
We commence angle chasing: we know $\angle DBC \cong DAC = \gamma$. Therefore \[\angle IBD = \alpha + \gamma\].
Looking at triangle $ABI$, we see that $\angle IBA = \alpha$, and $\angle BAI = \gamma$. Therefore because the sum of the angles must be $180$, $\angle BIA = 180-\alpha - \gamma$. Now $AD$ is a straight line, so \[\angle BID = 180-\angle BIA = \alpha+\gamma\].
Since $\angle IBD = \angle BID$, triangle $IBD$ is isosceles and thus $ID = BD$.
A similar argument should suffice to show $CD = ID$ by symmetry, so thus $ID = BD = CD$.
Now we regroup and get \[CD \cdot \frac{AL}{CL} \cdot BD \cdot \frac{AL}{BL} = ID^2 \cdot \frac{AL^2}{BL \cdot CL}\]
Now note that $BL$ and $CL$ are part of the same chord in the circle, so we can use Power of a point to express their product differently. \[BL \cdot CL = AL \cdot LD \implies AB \cdot AC = ID^2 \cdot \frac{AL}{LD}\]
Now we have some sort of expression for $AB \cdot AC$ in terms of $ID$ and $AL$. Let's try to find $AL$ first.
Drop an altitude from $D$ to $BC$, $I$ to $AC$, and $I$ to $BC$:

Since $\angle DBE \cong \angle IAF$ and $\angle BED \cong \angle IFA$, $\triangle BDE \sim \triangle AIF$.
Furthermore, we know $BD = ID$ and $AI = ID$, so $BD = AI$. Since we have two right similar triangles and the corresponding sides are equal, these two triangles are actually congruent: this implies that $DE = IF = 6$ since $IF$ is the inradius.
Now notice that $\triangle IGL \sim \triangle DEL$ because of equal vertical angles and right angles. Furthermore, $IG$ is the inradius so it's length is $6$, which equals the length of $DE$. Therefore these two triangles are congruent, so $IL = DL$.
Since $IL+DL = ID$, $ID = 2 \cdot IL$. Furthermore, $AL = AI + IL = ID + IL = 3 \cdot IL$.
We can now plug back into our initial equations for $AB \cdot AC$:
From $1.1$, $AB \cdot AC = 2 \cdot AL \cdot ID = 2 \cdot 3 \cdot IL \cdot 2 \cdot IL$
\[\implies AB \cdot AC = 3 \cdot (2 \cdot IL) \cdot (2 \cdot IL) = 3 \cdot ID^2\]
Alternatively, from $1.2$, $AB \cdot AC = ID^2 \cdot \frac{AL}{DL}$
\[\implies AB \cdot AC = ID^2  \cdot \frac{3 \cdot IL}{IL} = 3 \cdot ID^2\]
Now all we need to do is find $ID$.
The problem now becomes very simple if one knows Euler's Formula for the distance between the incenter and the circumcenter of a triangle. This formula states that $OI^2 = R(R-2r)$, where $R$ is the circumradius and $r$ is the inradius. We will prove this formula first, but if you already know the proof, skip this part.
Theorem: in any triangle, let $d$ be the distance from the circumcenter to the incenter of the triangle. Then $d^2 = R \cdot (R-2r)$, where $R$ is the circumradius of the triangle and $r$ is the inradius of the triangle.
Proof:
Construct the following diagram:

Let $OI = d$, $OH = R$, $IF = r$. By the Power of a Point, $IH \cdot IJ = AI \cdot ID$.
$IH = R+d$ and $IJ = R-d$, so \[(R+d) \cdot (R-d) = AI \cdot ID = AI \cdot CD\]
Now consider $\triangle ACD$. Since all three points lie on the circumcircle of $\triangle ABC$, the two triangles have the same circumcircle. Thus we can apply law of sines and we get $\frac{CD}{\sin(\angle DAC)} = 2R$. This implies
\[(R+d)\cdot (R-d) = AI \cdot 2R \cdot \sin(\angle DAC)\]
Also, $\sin(\angle DAC)) = \sin(\angle IAF))$, and $\triangle IAF$ is right. Therefore \[\sin(\angle IAF) = \frac{IF}{AI} = \frac{r}{AI}\]
Plugging in, we have
\[(R+d)\cdot (R-d) = AI \cdot 2R \cdot \frac{r}{AI} = 2R \cdot r\]
Thus \[R^2-d^2 = 2R \cdot r \implies d^2 = R \cdot (R-2r)\]

Now we can finish up our solution. We know that $AB \cdot AC = 3 \cdot ID^2$. Since $ID = AI$, $AB \cdot AC = 3 \cdot AI^2$. Since $\triangle AOI$ is right, we can apply the pythagorean theorem: $AI^2 = AO^2-OI^2 = 13^2-OI^2$.
Plugging in from Euler's formula, $OI^2 = 13 \cdot (13 - 2 \cdot 6) = 13$.
Thus $AI^2 = 169-13 = 156$.
Finally $AB \cdot AC = 3 \cdot AI^2 = 3 \cdot 156 = \textbf{468}$.

~KingRavi
By Euler's formula $OI^{2}=R(R-2r)$, we have $OI^{2}=13(13-12)=13$. Thus, by the Pythagorean theorem, $AI^{2}=13^{2}-13=156$. Let $AI\cap(ABC)=M$; notice $\triangle AOM$ is isosceles and $\overline{OI}\perp\overline{AM}$ which is enough to imply that $I$ is the midpoint of $\overline{AM}$, and $M$ itself is the midpoint of $II_{a}$ where $I_{a}$ is the $A$-excenter of $\triangle ABC$. Therefore, $AI=IM=MI_{a}=\sqrt{156}$ and \[AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.\]
Note that this problem is extremely similar to [2019 CIME I/14](https://artofproblemsolving.com/wiki/index.php/2019_CIME_I_Problems/Problem_14).
Denote $AB=a, AC=b, BC=c$. By the given condition, $\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6$, where $A$ is the area of $\triangle{ABC}$.
Moreover, since $OI\bot AI$, the second intersection of the line $AI$ and $(ABC)$ is the reflection of $A$ about $I$, denote that as $D$. By the incenter-excenter lemma, $DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c$.
Thus, we have $\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c$. Now, we have $\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}$
~Bluesoul
Denote by $R$ and $r$ the circumradius and inradius, respectively.
First, we have
\[
r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \hspace{1cm} (1)
\]
Second, because $AI \perp IO$,
\begin{align*}
AI & = AO \cos \angle IAO \\
& = AO \cos \left( 90^\circ - C - \frac{A}{2} \right) \\
& = AO \sin \left( C + \frac{A}{2} \right) \\
& = R \sin \left( C + \frac{180^\circ - B - C}{2} \right) \\
& = R \cos \frac{B - C}{2} .
\end{align*}
Thus,
\begin{align*}
r & = AI \sin \frac{A}{2} \\
& = R \sin \frac{A}{2} \cos \frac{B-C}{2} \hspace{1cm} (2)
\end{align*}
Taking $(1) - (2)$, we get
\[
4 \sin \frac{B}{2} \sin \frac{C}{2} = \cos \frac{B-C}{2} .
\]
We have
\begin{align*}
2 \sin \frac{B}{2} \sin \frac{C}{2}
& = - \cos \frac{B+C}{2} + \cos \frac{B-C}{2} .
\end{align*}
Plugging this into the above equation, we get
\[
\cos \frac{B-C}{2} = 2 \cos \frac{B+C}{2} . \hspace{1cm} (3)
\]
Now, we analyze Equation (2). We have
\begin{align*}
\frac{r}{R} & = \sin \frac{A}{2} \cos \frac{B-C}{2} \\
& = \sin \frac{180^\circ - B - C}{2} \cos \frac{B-C}{2} \\
& = \cos \frac{B+C}{2} \cos \frac{B-C}{2} \hspace{1cm} (4)
\end{align*}
Solving Equations (3) and (4), we get
\[
\cos \frac{B+C}{2} = \sqrt{\frac{r}{2R}}, \hspace{1cm}
\cos \frac{B-C}{2} = \sqrt{\frac{2r}{R}} . \hspace{1cm} (5)
\]
Now, we compute $AB \cdot AC$. We have
\begin{align*}
AB \cdot AC & = 2R \sin C \cdot 2R \sin B \\
& = 2 R^2 \left( - \cos \left( B + C \right) + \cos \left( B - C \right) \right) \\
& = 2 R^2 \left( - \left( 2 \left( \cos \frac{B+C}{2} \right)^2 - 1 \right)
+ \left( 2 \left( \cos \frac{B-C}{2} \right)^2 - 1 \right) \right) \\
& = 6 R r \\
& = \boxed{\textbf{(468) }}
\end{align*}
where the first equality follows from the law of sines, the fourth equality follows from (5).

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Answer: 468

<｜begin▁of▁sentence｜><｜User｜>Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$, circumradius $13$, and inradius $6$. Find $AB\cdot AC$.<｜Assistant｜><think>

Okay, so I have this geometry problem here: In triangle ABC, the circumcenter is O and the incenter is I. It's given that IA is perpendicular to OI. The circumradius is 13, and the inradius is 6. I need to find the product AB times AC. Hmm, okay, let me try to unpack this step by step.

First, let me recall some properties about triangle centers. The circumcenter O is the intersection of the perpendicular bisectors, and it's the center of the circumscribed circle with radius 13. The incenter I is where the angle bisectors meet, and it's the center of the inscribed circle with radius 6. The fact that IA is perpendicular to OI seems like a key condition here. I need to use that somehow.

Let me draw a rough sketch in my mind. Triangle ABC with O and I. IA is perpendicular to OI. Maybe I can place the triangle in coordinate system to make things easier? Sometimes coordinates help with such problems. Let me think about how to set that up.

Perhaps I can place point A at the origin? Wait, but then O and I would be somewhere else. Alternatively, maybe it's better to place the circumcenter O at the origin? That might simplify some calculations because the coordinates of A, B, C would then lie on a circle of radius 13. Hmm, but then the incenter I would be somewhere inside the triangle. The condition IA perpendicular to OI would mean that the vector from I to A is perpendicular to the vector from O to I. Maybe vectors could be useful here.

Alternatively, maybe using coordinate geometry with O at the origin. Let me try that.

Let me denote coordinates: Let’s set O at (0,0). Let’s let point A be somewhere on the circumcircle, so its coordinates can be (13,0) for simplicity, since the circumradius is 13. Wait, but then if I place A at (13,0), then the circumcenter is at (0,0), so OA is along the x-axis. Then, the incenter I is somewhere inside the triangle. The condition is that IA is perpendicular to OI. Let me see.

If O is at (0,0) and A is at (13,0), then vector OA is (13,0). The incenter I has coordinates (x,y). Then vector OI is (x,y), and vector IA is (13 - x, -y). The condition IA perpendicular to OI means their dot product is zero:

OI • IA = x*(13 - x) + y*(-y) = 13x - x² - y² = 0.

So that's one equation: 13x - x² - y² = 0. Hmm, that's equation (1).

Also, the inradius is 6. The inradius formula is r = Δ/s, where Δ is the area and s is the semiperimeter. So maybe I need to relate that. But to use that, I might need more information about the sides.

Alternatively, coordinates might get too messy here. Maybe there's a synthetic approach?

Wait, let me recall some relations between inradius, circumradius, and other elements. There's the formula involving the distance between incenter and circumcenter: OI² = R(R - 2r), where R is the circumradius and r is the inradius. Wait, is that true?

Wait, actually, the formula for the distance between the incenter and circumcenter is OI² = R² - 2Rr. Let me confirm that. Hmm, yes, I think that's a formula: OI² = R² - 2Rr cos A - ... Wait, maybe I need to check again.

Wait, actually, the formula is OI² = R² - 2Rr. Wait, no, perhaps that's only in certain cases. Let me recall correctly.

The formula for the distance between the incenter and circumcenter is given by:

OI² = R² - 2Rr

But I might be misremembering. Let me check the derivation.

Alternatively, the formula is OI² = R² - 2Rr, where R is the circumradius and r is the inradius. If that's the case, then here OI² = 13² - 2*13*6 = 169 - 156 = 13. So OI would be sqrt(13). But is this formula correct?

Wait, actually, I think the correct formula is OI² = R² - 2Rr. Let me confirm with a quick check. For an equilateral triangle, O and I coincide, so OI = 0. Then R = 2r in an equilateral triangle, since the centroid, circumcenter, and incenter all coincide. Wait, actually in an equilateral triangle, the inradius is (R/2). Wait, let me see:

In an equilateral triangle with side length a, the circumradius R = a/(√3), and the inradius r = (a√3)/6 = R/2. So then OI² would be 0, and R² - 2Rr = (a²/3) - 2*(a/√3)*(a/(2√3)) = (a²/3) - 2*(a²/(2*3)) = (a²/3 - a²/3) = 0. So that works. So maybe the formula is correct.

Therefore, in this problem, OI² = 13² - 2*13*6 = 169 - 156 = 13. Therefore, OI = sqrt(13). Hmm, okay, so that gives us the distance between O and I is sqrt(13). But in the problem, it's given that IA is perpendicular to OI. So maybe we can use some relations here.

Alternatively, perhaps using vector approaches. Let me think again.

Suppose we place the circumcenter O at the origin. Let me denote vectors OA, OB, OC as position vectors of points A, B, C. Since O is the circumcenter, |OA| = |OB| = |OC| = 13.

The incenter I has coordinates ( (aA + bB + cC)/ (a + b + c) ), where a, b, c are the lengths of the sides opposite to A, B, C respectively. Wait, actually, the incenter coordinates in barycentric coordinates are proportional to the lengths of the sides. But in Cartesian coordinates, if the triangle has vertices at A, B, C, then the incenter is ( (a x_A + b x_B + c x_C)/ (a + b + c), similarly for y-coordinates).

Hmm, this might get complicated, but maybe manageable.

Alternatively, since IA is perpendicular to OI, which is the vector from O to I. Let me denote vectors:

Let me denote vector OI as **v**, and vector IA as **w**. Then their dot product is zero: **v** • **w** = 0.

If O is at origin, then coordinates of I are (x, y), so vector OI is (x, y). Vector IA is A - I, so if A is at (13, 0), then IA is (13 - x, 0 - y). Then their dot product is x*(13 - x) + y*(-y) = 13x - x² - y² = 0, which is the same as equation (1) above.

Additionally, since O is the circumcenter, coordinates of A, B, C are all at distance 13 from O. Let me suppose that point A is at (13, 0). Then coordinates of B and C are somewhere on the circle of radius 13. Let me denote coordinates:

Let me set coordinate system with O at (0,0) and point A at (13,0). Then, coordinates of I are (x,y). Since I is the incenter, its coordinates can also be expressed in terms of the sides.

Let me denote the sides opposite to A, B, C as a, b, c respectively. Wait, standard notation: a = BC, b = AC, c = AB. Then the incenter coordinates would be ( (a x_A + b x_B + c x_C ) / (a + b + c), similarly for y).

Since point A is (13,0), and points B and C are somewhere on the circle x² + y² = 13². Let me denote coordinates of B as (13 cos β, 13 sin β) and C as (13 cos γ, 13 sin γ). Hmm, but this might involve too many variables. Maybe there's a smarter way.

Alternatively, since IA is perpendicular to OI, and we have that equation 13x - x² - y² = 0. Also, the inradius is 6. The inradius formula is r = Δ/s, where Δ is the area, and s is the semiperimeter.

So, if I can express Δ and s in terms of the sides, maybe I can relate them. But since we need AB * AC, which are sides c and b respectively (if a is BC, b is AC, c is AB). So the product is b*c. Hmm.

Alternatively, perhaps using trigonometric identities. Let me recall that in any triangle, the inradius r = 4R sin(A/2) sin(B/2) sin(C/2). Also, the distance OI is sqrt(R(R - 2r)). Wait, but we already used that OI² = R² - 2Rr, so OI = sqrt(13² - 2*13*6) = sqrt(169 - 156) = sqrt(13). So OI is sqrt(13). But how does that help?

Alternatively, since IA is perpendicular to OI, maybe we can relate the lengths IA and OI with some trigonometric relation.

Wait, IA is the distance from incenter I to vertex A. There is a formula for the distance from the incenter to a vertex. Let me recall that.

The distance from the incenter to vertex A is given by:

IA = \frac{r}{\sin \frac{A}{2}}

Similarly, IB = \frac{r}{\sin \frac{B}{2}}, IC = \frac{r}{\sin \frac{C}{2}}.

Is that correct? Let me verify.

Yes, in a triangle, the distance from the incenter to vertex A can be expressed as IA = \frac{r}{\sin \frac{A}{2}}. Because in the triangle, the inradius is related to the angles. Let me see:

In triangle AIF, where F is the point where the incircle touches side BC, then angle IAF is A/2, and the distance IA would be hypotenuse of a right triangle with angle A/2 and opposite side r. So sin(A/2) = r / IA => IA = r / sin(A/2). Yes, that seems right.

So IA = r / sin(A/2) = 6 / sin(A/2).

Also, the distance OI is sqrt(13). Since IA is perpendicular to OI, then in triangle OIA, which is a right triangle at A? Wait, no, IA is perpendicular to OI, so the triangle OIA has a right angle at I? Wait, let me think.

Wait, IA is perpendicular to OI. The vector IA is perpendicular to the vector OI. So the lines IA and OI are perpendicular. Therefore, the triangle OIA has a right angle at I. So triangle OIA is right-angled at I. Therefore, by Pythagoras:

OA² = OI² + IA².

Since OA is the circumradius, which is 13. So,

13² = (OI)² + (IA)²

We know OI is sqrt(13), so OI² = 13, so:

169 = 13 + IA² => IA² = 156 => IA = sqrt(156) = 2*sqrt(39).

But IA is also equal to 6 / sin(A/2), so:

6 / sin(A/2) = 2*sqrt(39)

Therefore,

sin(A/2) = 6 / (2*sqrt(39)) = 3 / sqrt(39) = sqrt(39)/13? Wait, let me compute:

Wait 3 divided by sqrt(39) can be rationalized as 3 sqrt(39)/39 = sqrt(39)/13. Yes, so sin(A/2) = sqrt(39)/13.

Therefore, sin(A/2) = sqrt(39)/13. Let me note that.

Now, perhaps we can find angle A from here. Let me compute cos(A/2) using sin² + cos² =1.

cos²(A/2) = 1 - (39/169) = (169 - 39)/169 = 130/169 => cos(A/2) = sqrt(130)/13.

Therefore, cos(A/2) = sqrt(130)/13.

Now, perhaps using the formula for the circumradius R = \frac{a}{2 \sin A} = 13. Wait, but a is the side BC. Hmm, but we might need to relate angle A with sides AB and AC.

Alternatively, let's recall that in triangle ABC, the inradius r = 6, and R =13. Also, angle A is related to sides b and c (AB and AC). Let me denote sides as follows:

Let me let BC = a, AC = b, AB = c. Then, the semiperimeter s = (a + b + c)/2.

The area Δ = r*s = 6s.

Also, the area can be expressed as Δ = (a*b*c)/(4R) = (a*b*c)/52.

Therefore, 6s = (a*b*c)/52 => (a*b*c) = 312 s. Hmm, but not sure if that helps yet.

Alternatively, using the formula for angle A:

In triangle ABC, angle A is between sides AB and AC, so the length BC can be expressed via the Law of Cosines:

a² = b² + c² - 2bc cos A.

Also, the area is (1/2)*b*c*sin A.

So, Δ = (1/2)*b*c*sin A = 6s.

Hmm, so (1/2)*b*c*sin A = 6s.

Also, from the formula R = \frac{a}{2 \sin A} => sin A = a/(2R) = a/26.

Therefore, sin A = a/26.

So, substituting back into the area equation:

(1/2)*b*c*(a/26) = 6s => (a b c)/52 = 6s => a b c = 312 s, which matches the earlier equation.

Hmm, so that's consistent but not helpful yet.

Alternatively, perhaps using the formula for cos(A/2). Since we have cos(A/2) = sqrt(130)/13, then cos A can be found via double angle formula:

cos A = 2 cos²(A/2) - 1 = 2*(130/169) -1 = 260/169 -1 = (260 - 169)/169 = 91/169 = 7*13 / 13² = 7/13.

Therefore, cos A = 7/13. Therefore, angle A has cosine 7/13, so sin A = sqrt(1 - (49/169)) = sqrt(120/169) = (2*sqrt(30))/13.

Therefore, sin A = 2 sqrt(30)/13.

So, from the Law of Sines, a = 2R sin A = 2*13*(2 sqrt(30)/13) = 4 sqrt(30). Wait, hold on:

Wait, Law of Sines says a / sin A = 2R, so a = 2R sin A. So:

a = 2*13*(2 sqrt(30)/13) = 2*2 sqrt(30) = 4 sqrt(30). Wait, that seems okay.

Wait, let me compute again:

sin A = 2 sqrt(30)/13, so:

a = 2 * 13 * (2 sqrt(30)/13) = 2 * 2 sqrt(30) = 4 sqrt(30). Yes, that's correct.

So side BC is a = 4√30.

Hmm, so now we know side a. Now, perhaps we can find other sides?

Alternatively, let's see if we can find the semiperimeter s and the other sides.

We also know that the inradius r = 6 = Δ/s, so Δ = 6s.

Also, the area Δ can be expressed as (1/2)*b*c*sin A. We have sin A = 2 sqrt(30)/13, so:

Δ = (1/2)*b*c*(2 sqrt(30)/13) = (b c sqrt(30))/13.

Therefore, (b c sqrt(30))/13 = 6s => b c = (6s *13)/sqrt(30) = (78 s)/sqrt(30).

Hmm, but not sure yet.

Alternatively, let's use the formula for the inradius:

r = (Δ)/s = 6, so Δ = 6s.

Also, from the formula involving the circumradius:

Δ = (a b c)/(4R) = (a b c)/52.

Therefore, 6s = (a b c)/52 => a b c = 312 s.

We already have a = 4√30, so:

4√30 * b * c = 312 s => b c = (312 s)/(4√30) = (78 s)/√30, which matches the previous equation. So that's consistent.

Hmm, so perhaps we need another equation to relate b and c.

We also know the semiperimeter s = (a + b + c)/2. So s = (4√30 + b + c)/2.

Therefore, s = 2√30 + (b + c)/2.

Hmm, so substituting s into the equation for bc:

b c = (78/(√30))*(2√30 + (b + c)/2 )

Wait, let me write that again:

b c = (78 s)/√30, and s = (a + b + c)/2 = (4√30 + b + c)/2.

Therefore,

b c = (78 / √30) * ( (4√30 + b + c)/2 )

Simplify:

Multiply 78 and 1/2: 39 / √30.

So,

b c = (39 / √30)*(4√30 + b + c )

Let me compute 39/√30 *4√30: 39 *4 * (√30 / √30) = 39*4 = 156.

Then, the rest is 39/(√30)*(b + c). So,

b c = 156 + (39/(√30))(b + c).

Hmm, this is getting a bit complicated. Maybe there's another relation.

Alternatively, perhaps using the formula for cos(A/2). Since we have cos(A/2) = sqrt(130)/13, and also, in triangle, cos(A/2) can be expressed in terms of the sides.

The formula for cos(A/2) is:

cos(A/2) = \sqrt{ \frac{(s)(s - a)}{b c} }.

Yes, that's a standard formula. Let me confirm:

Yes, in a triangle, cos(A/2) = sqrt( (s(s - a))/(b c) ). So,

cos²(A/2) = (s(s - a))/(b c).

We have cos²(A/2) = 130/169,

and s(s - a) / (b c) = 130/169.

So,

s(s - a) = (130/169) b c.

Hmm, let's plug in the known values.

We know a = 4√30,

so s - a = ( (a + b + c)/2 ) - a = (-a + b + c)/2 = ( -4√30 + b + c ) / 2.

Therefore,

s(s - a) = [ (4√30 + b + c)/2 ] * [ (-4√30 + b + c)/2 ]

= [ (b + c)^2 - (4√30)^2 ] / 4

= [ (b + c)^2 - 16*30 ] /4

= [ (b + c)^2 - 480 ] /4.

Therefore,

[ (b + c)^2 - 480 ] /4 = (130/169) b c.

Multiply both sides by 4:

(b + c)^2 - 480 = (520 / 169) b c.

Hmm, 520/169 simplifies to 520 divided by 13². 520 divided by 13 is 40, so 520/169 = 40/13.

Wait, 13*40 = 520, yes. So 520/169 = 40/13.

Therefore,

(b + c)^2 - 480 = (40/13) b c.

So,

(b + c)^2 = (40/13) b c + 480.

Hmm, okay, so that's one equation involving b and c.

Earlier, from the equation bc = 156 + (39/√30)(b + c). Let me see if I can use that.

Let me denote S = b + c and P = b c. Then, we have two equations:

1) S² = (40/13) P + 480,

2) P = 156 + (39/√30) S.

So substituting equation 2 into equation 1:

S² = (40/13)(156 + (39/√30) S ) + 480.

Let me compute this step by step.

First, compute (40/13)*156:

40/13 * 156 = 40 * (156 /13) = 40 * 12 = 480, since 156 divided by 13 is 12.

Then, (40/13)*(39/√30 S) = (40 * 39)/(13 √30) S = (40 * 3)/(√30) S = 120 / √30 S.

Therefore, equation becomes:

S² = 480 + (120 / √30) S + 480.

Wait, 480 (from first term) plus 480 (from the end) gives 960, so:

S² = 960 + (120 / √30) S.

Hmm, so:

S² - (120 / √30) S - 960 = 0.

This is a quadratic equation in S. Let me write it as:

S² - (120 / √30) S - 960 = 0.

Let me rationalize the coefficient:

120 / √30 = (120 √30)/30 = 4 √30.

Therefore, the equation becomes:

S² - 4√30 S - 960 = 0.

Let me solve for S using quadratic formula.

S = [4√30 ± sqrt( (4√30)^2 + 4*960 ) ] / 2.

Compute discriminant:

(4√30)^2 + 4*960 = 16*30 + 3840 = 480 + 3840 = 4320.

sqrt(4320) = sqrt(16*270) = 4*sqrt(270) = 4*sqrt(9*30) = 4*3*sqrt(30) = 12√30.

Therefore,

S = [4√30 ± 12√30]/2.

Since S = b + c is positive, we take the positive root:

S = (4√30 + 12√30)/2 = (16√30)/2 = 8√30.

Therefore, S = 8√30. So b + c = 8√30.

Then, from equation 2, P = 156 + (39/√30)*S.

Plugging S =8√30,

P = 156 + (39/√30)*(8√30) = 156 + 39*8 = 156 + 312 = 468.

Therefore, P = b c = 468. So the product AB * AC = b * c = 468. Wait, that seems like the answer? Let me confirm.

Wait, hold on. Let me check the steps again to make sure I didn't make a calculation error.

First, in the quadratic equation for S:

We had S² - 4√30 S - 960 = 0.

Discriminant: (4√30)^2 + 4*960 = 16*30 + 3840 = 480 + 3840 = 4320. sqrt(4320):

4320 = 16 * 270, so sqrt(16*270)=4*sqrt(270). Then 270=9*30, so sqrt(270)=3*sqrt(30). So sqrt(4320)=4*3*sqrt(30)=12√30. So that's correct.

Thus, S = [4√30 +12√30]/2 = 16√30 /2 =8√30. Correct.

Then, P = 156 + (39/√30)*(8√30). The term (39/√30)*(8√30) = 39*8*(√30 / √30) = 39*8 = 312. So 156 + 312 = 468. Correct.

Therefore, the product AB * AC = b*c = 468. So that's the answer?

Wait, but let me check if all steps are valid.

Wait, in the beginning, I assumed that O is at (0,0) and A is at (13,0). Then, through the condition IA perpendicular to OI, I derived that 13x -x² - y² =0. But then later, I switched to a trigonometric approach using formulas for IA and OI. So perhaps that coordinate system was not necessary, but the trigonometric approach worked out.

But let me confirm if the key step where I used the right triangle OIA is valid.

We were told IA is perpendicular to OI, so the lines IA and OI are perpendicular. Therefore, triangle OIA has a right angle at I. Therefore, OA² = OI² + IA². Since OA is the radius, 13², and OI we found via the formula OI² = R² - 2Rr = 169 - 156 =13. Therefore, OI is sqrt(13). Then IA squared is 169 -13 =156, so IA is sqrt(156). Then using IA = r / sin(A/2), so 6 / sin(A/2) = sqrt(156). Then sin(A/2)=6/sqrt(156)= 6/(2*sqrt(39))= 3/sqrt(39)= sqrt(39)/13 as above. Then proceeding to find cos(A/2), then cos A, etc. All steps seem okay.

Then using the Law of Sines to find side a = BC =4√30. Then setting up equations for semiperimeter and the product bc. Then solving the equations to get bc=468. So that seems consistent.

Alternatively, let me check with another approach to confirm.

Alternatively, using the formula for the distance from incenter to vertex:

IA = \frac{r}{\sin \frac{A}{2}} = \frac{6}{\sqrt{(1 - \cos A)/2}}.

Wait, since sin(A/2) = sqrt( (1 - cos A)/2 ). So IA = 6 / sqrt( (1 - cos A)/2 ) = 6 * sqrt(2/(1 - cos A)).

But we found cos A =7/13, so 1 - cos A = 6/13. Therefore,

IA = 6 * sqrt(2/(6/13)) = 6 * sqrt( (2*13)/6 ) = 6 * sqrt(26/6) = 6 * sqrt(13/3) = 6*(√13)/√3 = 2√13 * (√3)/√3 * Hmm, wait let me compute again:

Wait sqrt(26/6) = sqrt(13/3). So,

IA = 6 * sqrt(13/3) = 6*(sqrt(13)/sqrt(3)) = (6/sqrt(3)) * sqrt(13) = 2 sqrt(3) * sqrt(13) = 2 sqrt(39). Which matches the earlier result IA = sqrt(156) since sqrt(156)=sqrt(4*39)=2 sqrt(39). So that's consistent.

Therefore, all steps are consistent. Therefore, the product AB * AC is 468. So I think that is the answer.

Wait, but just to be thorough, let me check if with bc=468, and b + c =8√30, then the sides would satisfy the triangle inequality, etc.

Let me compute b and c. Suppose b and c are roots of the equation x² - Sx + P =0, so x² -8√30 x +468 =0.

Let me compute discriminant: (8√30)^2 -4*468 = 64*30 - 1872 = 1920 - 1872 =48.

Therefore, sqrt(48)=4√3. So roots are [8√30 ±4√3]/2 =4√30 ±2√3.

Therefore, sides b and c are 4√30 +2√3 and 4√30 -2√3. Both positive, since 4√30 is much larger than 2√3. So that's okay.

Then, the semiperimeter s = (a + b + c)/2 = (4√30 +8√30)/2 = (12√30)/2 =6√30.

Then, the area Δ =6s =6*6√30=36√30.

Alternatively, using the other area formula: (1/2)*b*c*sin A.

b*c=468, sin A=2 sqrt(30)/13.

So (1/2)*468*(2 sqrt(30)/13)= (468 /13)*sqrt(30)= 36 sqrt(30). Which matches. So that's correct.

Therefore, all the formulas check out. Therefore, the answer is 468.

**Final Answer**
\boxed{468}
</think>

Given triangle \( \triangle ABC \) with circumcenter \( O \) and incenter \( I \), where \( \overline{IA} \perp \overline{OI} \), circumradius \( R = 13 \), and inradius \( r = 6 \). We need to find \( AB \cdot AC \).

1. **Distance between Incenter and Circumcenter**:
   Using the formula \( OI^2 = R(R - 2r) \):
   \[
   OI^2 = 13^2 - 2 \cdot 13 \cdot 6 = 169 - 156 = 13 \implies OI = \sqrt{13}
   \]

2. **Right Triangle \( \triangle OIA \)**:
   Since \( \overline{IA} \perp \overline{OI} \), triangle \( \triangle OIA \) is a right triangle at \( I \):
   \[
   OA^2 = OI^2 + IA^2 \implies 13^2 = (\sqrt{13})^2 + IA^2 \implies IA^2 = 169 - 13 = 156 \implies IA = \sqrt{156}
   \]

3. **Angle \( A \) Calculation**:
   Using \( IA = \frac{r}{\sin \frac{A}{2}} \):
   \[
   \sqrt{156} = \frac{6}{\sin \frac{A}{2}} \implies \sin \frac{A}{2} = \frac{6}{\sqrt{156}} = \frac{\sqrt{39}}{13}
   \]
   \[
   \cos \frac{A}{2} = \sqrt{1 - \sin^2 \frac{A}{2}} = \sqrt{\frac{130}{169}} = \frac{\sqrt{130}}{13}
   \]
   \[
   \cos A = 2 \cos^2 \frac{A}{2} - 1 = \frac{7}{13}
   \]

4. **Side Lengths and Semiperimeter**:
   Using the Law of Sines:
   \[
   a = BC = 2R \sin A = 4\sqrt{30}
   \]
   Let \( b = AC \) and \( c = AB \). The semiperimeter \( s \):
   \[
   s = \frac{a + b + c}{2}
   \]
   Using the area formulas \( \Delta = r \cdot s \) and \( \Delta = \frac{1}{2} bc \sin A \):
   \[
   \Delta = 6s = \frac{1}{2} bc \cdot \frac{2\sqrt{30}}{13} \implies 6s = \frac{bc \sqrt{30}}{13}
   \]
   \[
   bc = \frac{78s}{\sqrt{30}}
   \]

5. **Solving for \( b \) and \( c \)**:
   Using the relationship \( \cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}} \):
   \[
   \frac{130}{169} = \frac{s(s - a)}{bc}
   \]
   Substituting \( s = \frac{4\sqrt{30} + b + c}{2} \) and solving the quadratic equations for \( b + c \) and \( bc \):
   \[
   b + c = 8\sqrt{30}, \quad bc = 468
   \]

Thus, the product \( AB \cdot AC = bc = \boxed{468} \).


Solution:
$a^2(b+c)+b^2(a+c)+c^2(a+b) = 6000000$, thus $a^2(300-a)+b^2(300-b)+c^2(300-c) = 6000000$. Complete the cube to get $-(a-100)^3-(b-100)^3+(c-100)^3 = 9000000-30000(a+b+c)$, which so happens to be 0. Then we have $(a-100)^3+(b-100)^3+(c-100)^3 = 0$. We can use Fermat's last theorem here to note that one of a, b, c has to be 100. We have 200+200+200+1 = 601.
We have
\begin{align*}
& a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \\
& = ab \left( a + b \right) + bc \left( b + c \right) + ca \left( c + a \right) \\
& = ab \left( 300 - c \right) + bc \left( 300 - a \right) + ca \left( 300 - b \right) \\
& = 300 \left( ab + bc + ca \right) - 3 abc \\
& = -3 \left(
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right)
- 10^4 \left( a + b + c \right) + 10^6
\right) \\
& = -3 \left(
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right)
- 2 \cdot  10^6
\right) \\
& = 6 \cdot 10^6 .
\end{align*}
The first and the fifth equalities follow from the condition that $a+b+c = 300$.
Therefore,
\[
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 .
\]
Case 1: Exactly one out of $a - 100$, $b - 100$, $c - 100$ is equal to 0.
Step 1: We choose which term is equal to 0. The number ways is 3.
Step 2: For the other two terms that are not 0, we count the number of feasible solutions.
W.L.O.G, we assume we choose $a - 100 = 0$ in Step 1. In this step, we determine $b$ and $c$.
Recall $a + b + c = 300$. Thus, $b + c = 200$.
Because $b$ and $c$ are nonnegative integers and $b - 100 \neq 0$ and $c - 100 \neq 0$, the number of solutions is 200.
Following from the rule of product, the number of solutions in this case is $3 \cdot 200 = 600$.
Case 2: At least two out of $a - 100$, $b - 100$, $c - 100$ are equal to 0.
Because $a + b + c = 300$, we must have $a = b = c = 100$.
Therefore, the number of solutions in this case is 1.
Putting all cases together, the total number of solutions is $600 + 1 = \boxed{\textbf{(601) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
We will use Vieta's formulas to solve this problem. We assume $a + b + c = 300$, $ab + bc + ca = m$, and $abc = n$. Thus $a$, $b$, $c$ are the three roots of a cubic polynomial $f(x)$.
We note that $300m = (a + b + c)(ab + bc + ca)=\sum_{cyc} a^2b + 3abc = 6000000 + 3n$, which simplifies to $100m - 2000000 = n$.
Our polynomial $f(x)$ is therefore equal to $x^3 - 300x^2 + mx - (100m - 2000000)$. Note that $f(100) = 0$, and by polynomial division we obtain $f(x) = (x - 100)(x^2 - 200x - (m-20000))$.
We now notice that the solutions to the quadratic equation above are $x = 100 \pm \frac{\sqrt{200^2 - 4(m - 20000)}}{2} = 100 \pm \sqrt{90000 - 4m}$, and that by changing the value of $m$ we can let the roots of the equation be any pair of two integers which sum to $200$. Thus any triple in the form $(100, 100 - x, 100 + x)$ where $x$ is an integer between $0$ and $100$ satisfies the conditions.
Now to count the possible solutions, we note that when $x \ne 100$, the three roots are distinct; thus there are $3! = 6$ ways to order the three roots. As we can choose $x$ from $0$ to $99$, there are $100 \cdot 3! = 600$ triples in this case. When $x = 100$, all three roots are equal to $100$, and there is only one triple in this case.
In total, there are thus $\boxed{601}$ distinct triples.
~GaloisTorrent <3
Let's define $a=100+x$, $b=100+y$, $c=100+z$. Then we have $x+y+z=0$ and $6000000 = \sum a^2(b+c)$
$= \sum (100+x)^2(200-x) = \sum (10000+200x+x^2)(200-x) = \sum (20000 - 10000 x + x(40000-x^2))$
$= \sum (20000 + 30000 x -x^3) = 6000000 - \sum x^3$, so we get $x^3 + y^3 + z^3 = 0$. Then from $x+y+z = 0$, we can find $0 = x^3+y^3+z^3 = x^3+y^3-(x+y)^3 = 3xyz$, which means that one of $a$, $b$,$c$ must be 0. There are 201 solutions for each of $a=0$, $b=0$ and $c=0$, and subtract the overcounting of 2 for solution $(200, 200, 200)$, the final result is $201 \times 3 - 2 = \boxed{601}$.
Dan Li

dan

Answer: 601

<｜begin▁of▁sentence｜><｜User｜>Find the number of triples of nonnegative integers \((a,b,c)\) satisfying \(a + b + c = 300\) and
\begin{equation*}
a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.
\end{equation*}<｜Assistant｜><think>

Okay, so I need to find the number of triples (a, b, c) of nonnegative integers such that a + b + c = 300 and the equation a²b + a²c + b²a + b²c + c²a + c²b = 6,000,000 holds. Hmm, let me start by understanding the problem step by step.

First, the variables a, b, c are nonnegative integers adding up to 300. The second equation looks a bit complicated, but maybe I can simplify it. Let me see if I can factor or rewrite that expression.

Looking at the second equation: a²b + a²c + b²a + b²c + c²a + c²b. Let me group terms:

a²b + a²c = a²(b + c)
Similarly, b²a + b²c = b²(a + c)
And c²a + c²b = c²(a + b)

So the entire expression becomes:

a²(b + c) + b²(a + c) + c²(a + b)

Hmm, that's a bit better. Since we know that a + b + c = 300, then (b + c) = 300 - a, (a + c) = 300 - b, and (a + b) = 300 - c. So substituting these into the expression:

a²(300 - a) + b²(300 - b) + c²(300 - c) = 6,000,000

So the equation simplifies to:

300(a² + b² + c²) - (a³ + b³ + c³) = 6,000,000

Hmm, that's a good simplification. Let me write that down:

300(a² + b² + c²) - (a³ + b³ + c³) = 6,000,000.

Now, maybe I can relate this to some known identities or find a way to express this in terms of symmetric sums?

Alternatively, perhaps I can consider variables in terms of each other. Since a + b + c = 300, maybe we can let two variables be expressed in terms of the third? But with three variables, that might get complicated. Alternatively, maybe assume some symmetry?

Alternatively, perhaps consider that the equation is symmetric in a, b, c, so maybe the solutions are symmetric in some way? Maybe the triples where two variables are equal, or all three are equal?

Wait, let me check if all three being equal is possible. If a = b = c, then each would be 100, since 100*3 = 300. Let me plug into the second equation:

Each term would be 100²*100 + ... but wait, let me compute the expression:

a²b + a²c + b²a + b²c + c²a + c²b. If a = b = c = 100, then each term is 100²*100, and there are 6 terms. Wait, actually, each pair:

Wait, actually, for a = b = c = 100,

Each term like a²b would be 100² * 100 = 1,000,000, and there are 6 such terms, so total would be 6,000,000. Hey, that's exactly the value we need! So (100, 100, 100) is a solution.

So that's one solution. But the problem is asking for the number of triples, so there might be more. Maybe permutations of this? But since all are equal, permutations would be the same. So that's just one triple.

But maybe there are other triples where variables are not all equal. Let me see.

Alternatively, perhaps two variables are equal, and the third is different. Let me suppose that a = b, then c = 300 - 2a.

Then, let's substitute into the equation:

First, let me write the equation again in terms of a, b, c:

300(a² + b² + c²) - (a³ + b³ + c³) = 6,000,000.

If a = b, then c = 300 - 2a. Let me substitute:

300(2a² + c²) - (2a³ + c³) = 6,000,000.

Substituting c = 300 - 2a:

First compute 2a² + c² = 2a² + (300 - 2a)^2 = 2a² + 90000 - 1200a + 4a² = 6a² - 1200a + 90000.

Similarly, 2a³ + c³ = 2a³ + (300 - 2a)^3.

Let me compute (300 - 2a)^3:

= 300³ - 3*300²*(2a) + 3*300*(2a)^2 - (2a)^3

= 27,000,000 - 3*90,000*2a + 3*300*4a² - 8a³

= 27,000,000 - 540,000a + 3600a² - 8a³

Therefore, 2a³ + (300 - 2a)^3 = 2a³ + 27,000,000 - 540,000a + 3600a² - 8a³

= (2a³ - 8a³) + 3600a² -540,000a +27,000,000

= -6a³ + 3600a² -540,000a +27,000,000

So putting back into the equation:

300*(6a² - 1200a + 90000) - [ -6a³ + 3600a² -540,000a +27,000,000 ] = 6,000,000

Let me compute each part step by step.

First, compute 300*(6a² - 1200a + 90000):

= 1800a² - 360,000a + 27,000,000

Then subtract the other part:

- [ -6a³ + 3600a² -540,000a +27,000,000 ]

= +6a³ - 3600a² +540,000a -27,000,000

So adding the two parts together:

1800a² - 360,000a + 27,000,000 + 6a³ - 3600a² +540,000a -27,000,000

Combine like terms:

6a³ + (1800a² - 3600a²) + (-360,000a + 540,000a) + (27,000,000 -27,000,000)

Simplify each term:

6a³ - 1800a² + 180,000a + 0

So the entire expression simplifies to:

6a³ - 1800a² + 180,000a = 6,000,000

Divide both sides by 6:

a³ - 300a² + 30,000a = 1,000,000

Bring all terms to left side:

a³ - 300a² + 30,000a - 1,000,000 = 0

Hmm, so we have a cubic equation in a. Let me see if I can factor this or find integer roots.

Possible rational roots are factors of 1,000,000 divided by factors of 1 (leading coefficient). So possible integer roots are divisors of 1,000,000. Since a is a nonnegative integer and since a must be less than or equal to 150 (since c = 300 - 2a must be nonnegative, so 300 - 2a ≥ 0 ⇒ a ≤ 150). So possible a is between 0 and 150. So possible roots in that range.

Let me try a = 100. Plugging into the equation:

100³ - 300*(100)^2 + 30,000*100 - 1,000,000

= 1,000,000 - 3,000,000 + 3,000,000 - 1,000,000 = 0. So a = 100 is a root.

Therefore, (a - 100) is a factor. Let's perform polynomial division or factor it.

Let me write the cubic as:

a³ - 300a² + 30,000a - 1,000,000 = (a - 100)(a² - 200a + 10,000)

Wait, let me check:

Multiply (a - 100)(a² - 200a + 10,000):

= a³ -200a² +10,000a -100a² +20,000a -1,000,000

= a³ - (200 + 100)a² + (10,000 + 20,000)a -1,000,000

= a³ -300a² +30,000a -1,000,000. Perfect, that's correct.

So the cubic factors as (a - 100)(a² - 200a + 10,000) = 0.

Now set each factor to zero:

First solution is a = 100.

Second quadratic equation: a² - 200a + 10,000 = 0.

Solving this quadratic:

a = [200 ± sqrt(200² - 4*1*10,000)] / 2

= [200 ± sqrt(40,000 - 40,000)] / 2

= [200 ± 0]/2 = 100.

So the quadratic has a double root at a = 100. Therefore, the cubic equation has a triple root at a = 100? Wait, no, the cubic factors as (a - 100)(a - 100)^2 = (a - 100)^3. Wait, but the quadratic was (a² - 200a + 10,000) which factors as (a - 100)^2. So indeed, the cubic is (a - 100)^3 = 0. So the only real root is a = 100, with multiplicity 3.

Therefore, the only solution in this case (when a = b) is a = b = 100, which gives c = 300 - 200 = 100. So that's the same solution as before, the triple (100, 100, 100). So in the case when a = b, the only solution is the all-100s.

Hmm, so maybe assuming two variables equal only gives the all-equal case. Let me see if there are other possibilities where variables are not equal.

Alternatively, perhaps one of the variables is zero? Let me check that case.

Suppose c = 0. Then the equations become a + b = 300, and the second equation becomes:

a²b + a²*0 + b²a + b²*0 + 0²a + 0²b = a²b + b²a = ab(a + b) = ab*300 = 6,000,000.

Therefore, ab*300 = 6,000,000 ⇒ ab = 20,000.

So we need nonnegative integers a, b with a + b = 300 and ab = 20,000.

So solving for a and b. Let me set up the quadratic equation. Let a be variable, then b = 300 - a.

Thus, a(300 - a) = 20,000 ⇒ 300a - a² = 20,000 ⇒ a² - 300a + 20,000 = 0.

Solving this quadratic:

a = [300 ± sqrt(90,000 - 80,000)] / 2 = [300 ± sqrt(10,000)] / 2 = [300 ± 100]/2.

Thus, a = (300 + 100)/2 = 400/2 = 200, or a = (300 - 100)/2 = 200/2 = 100.

Therefore, the solutions are (a, b, c) = (200, 100, 0) and (100, 200, 0). Since c is zero, but the problem allows nonnegative integers, so these are valid. However, since the variables are ordered triples, these are two distinct solutions. But also, since c could be any of the three variables, so if we set a = 0 or b = 0, similar solutions would exist.

Wait, so if c = 0 gives two solutions (200,100,0) and (100,200,0). Similarly, if a = 0, then we would have solutions (0, 200, 100) and (0, 100, 200). Similarly, if b = 0, then (200, 0, 100) and (100, 0, 200). So each of the three variables being zero gives two solutions, so total 6 solutions from these cases.

But wait, let me confirm. Let me check when a = 0:

If a = 0, then b + c = 300, and the equation becomes 0 + 0 + 0 + b²c + c²*0 + c²b = b²c + c²b = bc(b + c) = bc*300 = 6,000,000 ⇒ bc = 20,000. So same as before, leading to (0, 200, 100) and (0, 100, 200). Similarly for b = 0. So indeed, 6 solutions here.

Additionally, the all-100 case is another solution. So that's 7 so far. But maybe there are more?

Wait, perhaps there are solutions where none of the variables are zero and all are distinct? Let me see.

Alternatively, maybe there are solutions where two variables are zero? Let me check. Suppose two variables are zero, say b = c = 0. Then a = 300. Then plug into the second equation:

a²b + ... all terms would have at least one of b or c, so all terms are zero. So 0 ≠ 6,000,000. So that's invalid. Similarly, any two variables zero would give zero, which is too small. So no solutions there.

So so far, we have 7 solutions: the all-100s and six permutations of (200, 100, 0). But maybe there are more?

Wait, let me think again. Let me see if there are other triples where none of the variables are zero and not all equal.

Suppose that a, b, c are all positive integers, not all equal, and not having any zeros. Let me see if such triples can satisfy the equations.

Alternatively, perhaps the only solutions are the ones we found. Let me see.

Alternatively, maybe there are solutions where two variables are equal, but not leading to the all-equal case. Wait, earlier when I assumed a = b, the only solution was a = b = c = 100. So perhaps that's the only case when two variables are equal.

Alternatively, maybe there are solutions where variables are in some proportion?

Alternatively, perhaps I can consider the original equation again. Let me recall that we had:

300(a² + b² + c²) - (a³ + b³ + c³) = 6,000,000.

Alternatively, maybe I can write this as:

a³ + b³ + c³ = 300(a² + b² + c²) - 6,000,000.

Hmm, not sure if that helps. Alternatively, perhaps using the identity for a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca). Maybe that can be useful here?

Let me see. Let me denote S = a + b + c = 300,

Q = a² + b² + c²,

P = abc,

and let me denote the sum ab + bc + ca = M.

We know that Q = S² - 2M = 300² - 2M = 90,000 - 2M.

Also, the expression a³ + b³ + c³ can be written as S³ - 3S M + 3P.

Therefore, substituting into the equation:

300 Q - (a³ + b³ + c³) = 6,000,000

Substitute Q and a³ + b³ + c³:

300*(90,000 - 2M) - [300³ - 3*300*M + 3P] = 6,000,000

Let me compute each term step by step.

First, compute 300*Q = 300*(90,000 - 2M) = 27,000,000 - 600M.

Then compute the cubic term:

300³ = 27,000,000,

so the cubic expression is 27,000,000 - 3*300*M + 3P = 27,000,000 - 900M + 3P.

Therefore, the entire left-hand side becomes:

[27,000,000 - 600M] - [27,000,000 - 900M + 3P] =

27,000,000 - 600M -27,000,000 + 900M - 3P =

( -600M + 900M ) + (27,000,000 -27,000,000) -3P =

300M - 3P = 6,000,000.

Therefore, the equation simplifies to:

300M - 3P = 6,000,000 ⇒ divide both sides by 3:

100M - P = 2,000,000 ⇒ P = 100M - 2,000,000.

So we have that abc = 100(ab + bc + ca) - 2,000,000.

Hmm, so now we have a relation between abc and ab + bc + ca. Maybe this can help?

We also know that a + b + c = 300, so perhaps we can express M in terms of variables?

Alternatively, perhaps we can consider variables in terms of two variables?

Alternatively, let me see if the solutions we found satisfy this equation.

First, the all-100 case:

a = b = c = 100.

Then ab + bc + ca = 3*(100*100) = 30,000,

abc = 100^3 = 1,000,000.

Then 100M - P = 100*30,000 - 1,000,000 = 3,000,000 - 1,000,000 = 2,000,000. Which matches.

Good, so that works.

Now, take one of the other solutions, say (200, 100, 0).

Compute ab + bc + ca: 200*100 + 100*0 + 0*200 = 20,000 + 0 + 0 = 20,000.

abc = 200*100*0 = 0.

Then 100M - P = 100*20,000 - 0 = 2,000,000. Which also matches. So that works too.

So this equation holds for the found solutions. So perhaps all solutions must satisfy this equation. So maybe we can use this equation to find other solutions?

Alternatively, perhaps we can parametrize variables?

Alternatively, let me think of variables as follows. Let me suppose that one variable is 0, which gives the solutions we found. If none are zero, then perhaps the variables are all positive. Let me see if there are solutions with all variables positive and distinct.

Suppose that a, b, c are all positive integers, not all equal, and none zero. Let me see if such triples can exist.

Alternatively, perhaps we can consider the equation P = 100M - 2,000,000. Since P = abc, so abc must equal 100(ab + bc + ca) - 2,000,000.

Hmm, perhaps rearranged as abc - 100(ab + bc + ca) = -2,000,000.

Alternatively, factor:

abc - 100(ab + bc + ca) = -2,000,000.

Hmm, not sure. Alternatively, perhaps factor terms:

Let me factor terms with ab, bc, ca:

ab(c - 100) + bc(a - 100) + ca(b - 100) = ?

Wait, maybe not straightforward. Alternatively, perhaps adding 100^3 to both sides?

Wait, let me see:

abc - 100(ab + bc + ca) + 100^3 = 100^3 - 2,000,000.

Wait, 100^3 is 1,000,000, so RHS would be 1,000,000 - 2,000,000 = -1,000,000. Not sure.

Alternatively, perhaps think of (a - 100)(b - 100)(c - 100). Let me compute that:

(a - 100)(b - 100)(c - 100) = abc - 100(ab + bc + ca) + 100^2(a + b + c) - 100^3.

From our previous equation, abc - 100(ab + bc + ca) = -2,000,000. So substituting:

= (-2,000,000) + 100^2*(300) - 100^3

Compute 100^2 * 300 = 10,000 * 300 = 3,000,000

100^3 = 1,000,000

Thus,

= -2,000,000 + 3,000,000 - 1,000,000 = 0.

Therefore, (a - 100)(b - 100)(c - 100) = 0.

Ah! That's a key insight. So this product equals zero, which implies that at least one of the terms must be zero. Therefore, at least one of a, b, c must be equal to 100.

Therefore, all solutions must have at least one variable equal to 100. That's a crucial point.

Therefore, all triples (a, b, c) satisfying the given equations must have at least one of a, b, c equal to 100. So this reduces the problem to considering cases where one variable is 100, and the other two variables sum to 200, and so on.

Therefore, let me structure the solution based on this.

Case 1: Exactly one of a, b, c is 100.

Case 2: Two of them are 100.

Case 3: All three are 100.

We already know that the all-three case is valid. Let's see.

Case 3: All three are 100: (100, 100, 100). That's one solution.

Case 2: Two of them are 100. Let's say a = 100, b = 100, then c = 100. Wait, that's the same as case 3. So if two are 100, then the third must also be 100, since a + b + c = 300. Wait, 100 + 100 + c = 300 ⇒ c = 100. So actually, if two variables are 100, the third must be 100 as well. Therefore, there are no solutions with exactly two variables equal to 100. So case 2 doesn't contribute any new solutions.

Therefore, the only possibilities are Case 1 (exactly one variable is 100) and Case 3 (all three variables are 100).

So let's analyze Case 1: Exactly one of a, b, c is 100. Without loss of generality, let's assume a = 100. Then b + c = 200. We need to find nonnegative integers b and c such that when a = 100, the second equation is satisfied.

Let me substitute a = 100 into the equation:

The original equation is:

a²b + a²c + b²a + b²c + c²a + c²b = 6,000,000.

Plugging a = 100:

100²b + 100²c + b²*100 + b²c + c²*100 + c²b = 6,000,000.

Let me compute each term:

100²(b + c) = 10,000*(200) = 2,000,000 (since b + c = 200)

Then the remaining terms:

b²*100 + b²c + c²*100 + c²b.

Let me factor terms:

= 100(b² + c²) + bc(b + c)

Because:

b²c + c²b = bc(b + c),

and the other terms are 100b² + 100c².

Therefore, total expression becomes:

2,000,000 + 100(b² + c²) + bc*(200) = 6,000,000.

Thus,

100(b² + c²) + 200bc = 6,000,000 - 2,000,000 = 4,000,000.

Divide both sides by 100:

b² + c² + 2bc = 40,000.

Wait, but b² + c² + 2bc = (b + c)^2 = 200² = 40,000. So that's exactly equal to 40,000. Therefore, the equation reduces to 40,000 = 40,000, which is always true.

Wait a second! That means that whenever a = 100 and b + c = 200, the equation is automatically satisfied?

Hmm, so that suggests that all triples where one variable is 100 and the other two sum to 200 will satisfy the equation. But that can't be right because when we considered the case where one variable is 0, like (200, 100, 0), that worked, but what about other triples?

Wait, let me check with another example. Suppose a = 100, b = 150, c = 50. Then let's compute the second equation:

First, a + b + c = 300, which is okay.

Compute the expression:

a²b + a²c + b²a + b²c + c²a + c²b.

Compute each term:

a²b = 100²*150 = 10,000*150 = 1,500,000

a²c = 100²*50 = 10,000*50 = 500,000

b²a = 150²*100 = 22,500*100 = 2,250,000

b²c = 150²*50 = 22,500*50 = 1,125,000

c²a = 50²*100 = 2,500*100 = 250,000

c²b = 50²*150 = 2,500*150 = 375,000

Adding all these up:

1,500,000 + 500,000 = 2,000,000

2,250,000 + 1,125,000 = 3,375,000

250,000 + 375,000 = 625,000

Total: 2,000,000 + 3,375,000 + 625,000 = 6,000,000. Hey, that works!

So indeed, any triple where one variable is 100 and the other two sum to 200 will satisfy the equation. So that means there are infinitely many solutions? Wait, but the problem states nonnegative integers, so the number of triples would be the number of nonnegative integer solutions to b + c = 200, multiplied by 3 (for each variable being the 100), minus overlaps?

Wait, but hold on. Wait, the problem says nonnegative integers, so for each case where one variable is fixed as 100, the other two variables can be any pair (b, c) with b + c = 200. The number of nonnegative integer solutions to b + c = 200 is 201 (from b=0 to b=200). Since there are three variables, each can be the one set to 100, so 3 * 201. However, we have to subtract overlaps where two variables are 100, but as we saw earlier, if two variables are 100, then the third must also be 100, so the only overlap is the triple (100,100,100), which is counted three times in the 3*201 count (once for each variable being the fixed 100). Therefore, total number would be 3*201 - 3*1 + 1? Wait, let me think carefully.

Wait, actually, when we count for each variable being 100, the case where all three are 100 is included in each of the three counts. So the total count would be 3*(201) - 3*(1) + 1? Wait, maybe inclusion-exclusion.

Alternatively, let me compute it step by step.

First, the number of triples where a = 100: then b + c = 200, which has 201 solutions (b from 0 to 200, c = 200 - b).

Similarly for b = 100: a + c = 200, 201 solutions.

Similarly for c = 100: a + b = 200, 201 solutions.

Total so far: 3*201 = 603.

However, the triple (100, 100, 100) is counted three times, once in each of the three cases. But it should be counted only once. So we need to subtract the duplicates.

The overcount is 2 (since it's counted 3 times but should be 1, so subtract 2). Alternatively, using inclusion-exclusion:

Total = (number with a=100) + (number with b=100) + (number with c=100) - (number with a=100 and b=100) - (number with a=100 and c=100) - (number with b=100 and c=100) + (number with all three 100).

Now, the intersections where two variables are 100: For example, a=100 and b=100 implies c=100, so only one solution each. There are three such intersections, each contributing 1. The intersection of all three is also 1.

Therefore:

Total = 201 + 201 + 201 - 1 -1 -1 +1 = 603 - 3 +1 = 601.

Wait, let me compute:

201*3 = 603

Subtract 3 (for the three pairwise intersections each contributing 1): 603 - 3 = 600

Add back 1 for the triple intersection: 600 +1 = 601.

Therefore, total number of triples is 601?

But wait, but earlier we saw that when one variable is 100 and the other two sum to 200, all those triples satisfy the equation. So that would include the all-100 case as well. So the total number is 601?

But hold on, let me verify with an example. Let me see:

Take the case where a = 100, b = 0, c = 200. That is one of the solutions we found earlier, and it's included in the count for a=100. Similarly, (100, 200, 0) is also in there. The all-100 is counted three times but adjusted to once.

However, the problem is asking for triples of nonnegative integers. So this seems correct. But wait, let me check another case where a = 100, b = 150, c = 50, which we tested earlier, that works and is counted here.

Therefore, according to this, the total number of triples is 601. But wait, but earlier when we considered the case where one variable is zero, like (200, 100, 0), those are included in this count. Because when a = 100, b can be 200 and c = 0, which is allowed. So all those 6 permutations are included here?

Wait, actually, when we fix a variable to 100, the other two can be anything as long as they sum to 200, including cases where one is zero. So the triples where one variable is 100, another is 200, and the last is 0 are included here. For example, (100, 200, 0) is part of the a=100 case (with b=200, c=0). Similarly, (200, 100, 0) is part of the b=100 case (with a=200, c=0). So all those permutations are accounted for in the 601 count.

But wait, let me compute 201*3 - 3 +1 again. Wait, 201*3 is 603, subtract 3 gives 600, plus 1 is 601. Hmm.

But let me think of it another way. The number of triples where at least one variable is 100 is equal to the total number of triples with a=100, plus those with b=100, plus those with c=100, minus overlaps. Which is exactly what we did.

Alternatively, the formula for the number of non-negative integer solutions with at least one variable equal to 100 is:

For each variable fixed at 100, the other two sum to 200, so 201 each, so 3*201. But subtract the overlaps where two variables are 100, which as we saw only occurs when all three are 100, so three cases where two variables are fixed at 100, each allowing only one solution (the third variable 100). So subtract 3*(1) but then add back the one case where all three are 100 because it was subtracted too many times. So total is 3*201 - 3*1 + 1 = 603 -3 +1 = 601.

Therefore, the total number of triples is 601. But wait, but earlier I thought there were 7 solutions, but that was only considering the cases where variables are 0 or 100 or 200. But actually, there are many more solutions where the variables are other numbers, like (100, 150, 50), etc. So the correct total is 601?

But let me check with another example. Let me pick a = 100, b = 100, c = 100. That is counted once. Then, for a = 100, b from 0 to 200 gives 201 triples. Similarly for the others. So 601 seems correct.

Wait, but the problem says "nonnegative integers", so zeros are allowed. So this seems okay.

But let me check if there are any other solutions outside of these cases. Earlier, we had the equation (a - 100)(b - 100)(c - 100) = 0, which means that at least one variable must be 100. Therefore, all solutions must have at least one variable equal to 100. Therefore, all solutions are counted in the 601 total. Therefore, the answer should be 601?

Wait, but let me confirm with another example. Let me pick a = 100, b = 1, c = 199. Then the sum is 100 + 1 + 199 = 300. Let's check the second equation:

Compute the expression:

a²b + a²c + b²a + b²c + c²a + c²b.

Compute each term:

a²b = 10000 * 1 = 10,000

a²c = 10000 * 199 = 1,990,000

b²a = 1 * 100 = 100

b²c = 1 * 199 = 199

c²a = (199)^2 * 100. Let's compute 199² = (200 -1)^2 = 40,000 - 400 + 1 = 39,601. So 39,601 * 100 = 3,960,100

c²b = 39,601 * 1 = 39,601

Adding all terms:

10,000 + 1,990,000 = 2,000,000

100 + 199 = 299

3,960,100 + 39,601 = 4,000, (wait 3,960,100 + 39,601 = 3,960,100 + 39,601 = 3,999,701?)

Wait let me compute step by step:

First terms: a²b + a²c = 10,000 + 1,990,000 = 2,000,000

Next terms: b²a + b²c = 100 + 199 = 299

Next terms: c²a + c²b = 3,960,100 + 39,601 = 3,999,701

Total sum: 2,000,000 + 299 + 3,999,701 = 2,000,000 + 3,999,701 = 5,999,701 + 299 = 6,000,000. Perfect, so that works.

Therefore, indeed, all triples with at least one 100 and the other two summing to 200 work. So the total number is 601.

Wait, but let me check the count again. Let me think of variables:

For each variable being 100, the other two variables can be any pair (x, y) such that x + y = 200. The number of non-negative integer solutions for x + y = 200 is 201 (from x=0 to x=200). So for each of the three variables, 201 solutions. However, when two variables are 100, like (100, 100, 100), it is counted in all three counts. So total count is 3*201 - 3*(number of solutions where two variables are 100) + number where all three are 100.

Wait, the number of solutions where two variables are 100: For example, a=100 and b=100 requires c=100, so only one solution. Similarly for a=100 and c=100, etc. So there are three such solutions where two variables are fixed at 100, but each actually corresponds to the same triple (100,100,100). So the overlaps between the counts are each the single triple (100,100,100). Therefore, using inclusion-exclusion:

Total = (A + B + C) - (AB + AC + BC) + ABC,

where A is the count for a=100, B for b=100, C for c=100,

AB is the count for a=100 and b=100, which is 1,

AC is a=100 and c=100, which is 1,

BC is b=100 and c=100, which is 1,

and ABC is all three, which is 1.

Therefore,

Total = (201 + 201 + 201) - (1 +1 +1) + 1 = 603 - 3 +1 = 601. So that's correct.

Therefore, the total number of triples is 601.

Wait, but the problem says "triples of nonnegative integers (a, b, c)". So ordered triples? Yes, I think so. Because in the problem statement, it's asking for triples (a, b, c), so ordered. Therefore, the count of 601 is correct.

But let me check with the earlier examples:

The triple (100, 100, 100) is counted once.

The triples where one variable is 0, like (200, 100, 0), (100, 200, 0), etc., each of these is a distinct ordered triple. For example, (200, 100, 0) is part of the count when b=100 (since in that case, a + c = 200, so a=200, c=0). Similarly, (0, 200, 100) is part of c=100's count. So all permutations are included in the 601.

Therefore, the answer should be 601. But let me check if there is any mistake here.

Wait, but in the problem statement, the equation is a²b + a²c + ... = 6,000,000. We saw that when one variable is 100 and the others sum to 200, this equation holds. And the equation (a - 100)(b - 100)(c - 100) = 0 is necessary, so all solutions must have at least one variable 100. Therefore, all solutions are counted in the 601. So I think that is correct.

Wait, but let me check if there is an alternative way to compute this.

Alternatively, since each solution must have at least one 100, the total number is equal to the number of triples with a=100, plus those with b=100, plus those with c=100, minus overlaps, which we did.

Alternatively, the number of triples where a=100 is 201, same for b and c. So 3*201 - 3*1 +1 = 601.

Alternatively, think of it as for each position, the number of triples where that position is 100 and the other two sum to 200, so 201 each, but subtract duplicates.

Alternatively, the formula for the number of non-negative integer solutions to a + b + c = 300 with at least one variable equal to 100 is equal to:

Total = (number with a=100) + (number with b=100) + (number with c=100) - (number with a=100 and b=100) - ... + (number with all three 100).

Which is exactly inclusion-exclusion, so 601.

Therefore, I think the answer is 601.

Wait, but let me check with a smaller case to see if the method holds.

Suppose instead of 300 and 6,000,000, let's take a smaller number. Let me see.

Suppose we have a + b + c = 6, and the equation a²b + a²c + b²a + b²c + c²a + c²b = some value. Let's see if the same logic applies.

Suppose we set variables such that one of them is 2 (since 6/3 = 2). Let me see.

Suppose the equation is similar, scaled down. Let me see if the product (a - 2)(b - 2)(c - 2) = 0.

Wait, maybe this approach is consistent.

Alternatively, perhaps the answer is 601. But let me see, in the original problem, the user might expect the answer to be 601, but maybe I made a miscalculation in the inclusion-exclusion.

Wait, let me recalculate:

Number of triples with a=100: 201

Similarly for b and c. So 3×201 = 603.

Now, subtract the triples where two variables are 100. For example, triples where a=100 and b=100: then c must be 100, so only one triple: (100,100,100). Similarly for a=100 and c=100, and b=100 and c=100. So three such triples, but each is the same triple (100,100,100). So we have overcounted this triple three times in the initial 603, so we need to subtract 2 of them to leave one. Wait, inclusion-exclusion says subtract the intersections, which are each the single triple. Since there are three intersections (a=100 and b=100, etc.), each contributing 1, so subtract 3×1 = 3.

Then add back the triple where all three are 100, because it was subtracted too many times. Wait, in inclusion-exclusion, the formula is:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|.

Here, the total number we want is exactly |A ∪ B ∪ C|, which is the number of triples where at least one variable is 100. So:

|A| = |B| = |C| = 201,

|A ∩ B| = |A ∩ C| = |B ∩ C| = 1 (only the triple where all are 100),

|A ∩ B ∩ C| = 1.

Therefore,

|A ∪ B ∪ C| = 201×3 - 1×3 + 1 = 603 - 3 +1 = 601.

Therefore, correct.

Therefore, the answer is 601.

But wait, the problem says "nonnegative integers", so zeros are allowed, which we have considered. Therefore, I think the correct answer is 601.

However, let me check once more with another example. Let me take a case where two variables are 100, but that forces the third to be 100, so no other cases. So all solutions are either with exactly one 100 and the other two summing to 200, or all three 100. The count for exactly one 100 is 3*(200) because for each variable set to 100, there are 200 solutions where the other two are non-100? Wait no, actually when you fix a=100, the other two variables can be anything from 0 to 200 as long as they sum to 200. So the number of solutions where exactly one variable is 100 would be for each variable:

For a=100, the number of solutions where b and c are not 100. Wait, but if b or c is 100, then that would mean two variables are 100, which only occurs when all three are 100.

Wait, actually, if a=100, then to have exactly one variable as 100, we need b + c = 200 with neither b nor c equal to 100. The number of such solutions is 201 (total) minus 1 (the case where b=100 and c=100). So 200. So for each variable, the number of triples where exactly that variable is 100 and the others are not is 200. So total for exactly one variable 100 is 3×200 = 600. Then add the one case where all three are 100, totaling 601. That also matches.

Therefore, this cross-verification also gives 601. So I think that's the correct answer.

Therefore, the number of triples is \boxed{601}.

**Final Answer**
\boxed{601}
</think>

To find the number of triples \((a, b, c)\) of nonnegative integers satisfying \(a + b + c = 300\) and \(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000\), we start by simplifying the given equation.

First, we rewrite the second equation using the identity:
\[
a^2(b + c) + b^2(a + c) + c^2(a + b)
\]
Since \(a + b + c = 300\), we substitute \(b + c = 300 - a\), \(a + c = 300 - b\), and \(a + b = 300 - c\) into the equation:
\[
a^2(300 - a) + b^2(300 - b) + c^2(300 - c) = 6,000,000
\]
This simplifies to:
\[
300(a^2 + b^2 + c^2) - (a^3 + b^3 + c^3) = 6,000,000
\]

Using the identity for the sum of cubes and symmetric sums, we find that:
\[
(a - 100)(b - 100)(c - 100) = 0
\]
This implies that at least one of \(a, b, c\) must be 100.

We consider the cases where one of the variables is 100. Without loss of generality, let \(a = 100\). Then \(b + c = 200\). The number of nonnegative integer solutions to \(b + c = 200\) is 201. Similarly, if \(b = 100\) or \(c = 100\), we also get 201 solutions each.

Using inclusion-exclusion principle to account for overlaps:
- The total number of solutions when any one variable is 100 is \(3 \times 201\).
- Subtract the overcounted cases where two variables are 100 (each such case only allows the third variable to be 100, giving 1 solution each, totaling 3).
- Add back the case where all three variables are 100, which was subtracted three times.

Thus, the calculation is:
\[
3 \times 201 - 3 \times 1 + 1 = 603 - 3 + 1 = 601
\]

Therefore, the number of triples \((a, b, c)\) is \(\boxed{601}\).