Solution:
$\frac{9}{s} + t = 4$ in hours and $\frac{9}{s+2} + t = 2.4$ in hours.
Subtracting the second equation from the first, we get,
$\frac{9}{s} - \frac{9}{s+2} = 1.6$
Multiplying by $(s)(s+2)$, we get
$9s+18-9s=18=1.6s^{2} + 3.2s$
Multiplying by 5/2 on both sides, we get
$0 = 4s^{2} + 8s - 45$
Factoring gives us
$(2s-5)(2s+9) = 0$, of which the solution we want is $s=2.5$.
Substituting this back to the first equation, we can find that $t = 0.4$ hours.
Lastly, $s + \frac{1}{2} = 3$ kilometers per hour, so
$\frac{9}{3} + 0.4 = 3.4$ hours, or $\framebox{204}$ minutes
-Failure.net
The amount of hours spent while walking on the first travel is $\frac{240-t}{6}$. Thus, we have the equation $(240-t)(s) = 540$, and by the same logic, the second equation yields $(144-t)(s+2) = 540$. We have $240s-st = 540$, and $288+144s-2t-st = 540$. We subtract the two equations to get $96s+2t-288 = 0$, so we have $48s+t = 144$, so $t = 144-48s$, and now we have $(96+48s)(s) = 540$. The numerator of $s$ must evenly divide 540, however, $s$ must be less than 3. We can guess that $s = 2.5$. Now, $2.5+0.5 = 3$. Taking $\frac{9}{3} = 3$, we find that it will take three hours for the 9 kilometers to be traveled. The t minutes spent at the coffeeshop can be written as $144-48(2.5)$, so t = 24. $180 + 24 = 204$. -sepehr2010

Answer: 204

<｜begin▁of▁sentence｜><｜User｜>Every morning Aya goes for a $9$-kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of $s$ kilometers per hour, the walk takes her 4 hours, including $t$ minutes spent in the coffee shop. When she walks $s+2$ kilometers per hour, the walk takes her 2 hours and 24 minutes, including $t$ minutes spent in the coffee shop. Suppose Aya walks at $s+\frac{1}{2}$ kilometers per hour. Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.<｜Assistant｜><think>

Okay, let's see. I need to solve this problem about Aya's walking and coffee shop stops. Hmm. Let me read it again carefully.

Every morning Aya goes for a 9-kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of s kilometers per hour, the walk takes her 4 hours, including t minutes spent in the coffee shop. When she walks s+2 kilometers per hour, the walk takes her 2 hours and 24 minutes, including t minutes spent in the coffee shop. Suppose Aya walks at s+1/2 kilometers per hour. Find the number of minutes the walk takes her, including the t minutes spent in the coffee shop.

Alright, so the key points here are:

1. The distance of the walk is 9 kilometers.
2. When her speed is s km/h, the total time taken (including coffee break) is 4 hours. The coffee break is t minutes.
3. When her speed is s+2 km/h, the total time is 2 hours and 24 minutes, which is 2.4 hours? Wait, 24 minutes is 0.4 hours? Let me check: 24 divided by 60 is 0.4, yes. So 2.4 hours total, again including the same t minutes.
4. We need to find the total time when she walks at s + 1/2 km/h, again including the t minutes.

So, the total time includes both the walking time and the coffee shop time. The coffee shop time is the same in all cases, right? So t minutes is fixed. So, the difference in total time between the two scenarios must be due to the difference in walking time.

Let me denote variables more precisely.

Let me let T1 be the total time when speed is s: T1 = 4 hours = 240 minutes. This includes t minutes at the coffee shop, so the walking time is (240 - t) minutes.

Similarly, when the speed is s + 2 km/h, the total time T2 is 2 hours 24 minutes, which is 144 minutes. So the walking time here is (144 - t) minutes.

The distance is the same, 9 km, so we can write equations based on speed = distance / time. But note that time here is in hours, so need to convert walking time to hours.

Wait, let me think again. Let me define variables properly.

Let me denote:

Let t be the time spent in the coffee shop, in hours. Wait, but the problem says t is in minutes. Hmm, the problem says "t minutes spent in the coffee shop". So, the total time is walking time plus t minutes. Since the total time is given in hours and minutes, perhaps I need to convert everything to hours or everything to minutes. Maybe minutes would be better here.

Let me try to set up equations in minutes.

First scenario:

Speed = s km/h. Total time = 4 hours = 240 minutes. The coffee shop time is t minutes, so the walking time is (240 - t) minutes. But walking time in hours is (240 - t)/60 hours. The distance is 9 km, so:

Distance = speed × time. So,

9 = s × [(240 - t)/60]

Similarly, second scenario:

Speed = s + 2 km/h. Total time = 2 hours 24 minutes = 144 minutes. So walking time is (144 - t) minutes, which is (144 - t)/60 hours. Then,

9 = (s + 2) × [(144 - t)/60]

So now we have two equations:

1) 9 = s * (240 - t)/60

2) 9 = (s + 2) * (144 - t)/60

We can solve these two equations for s and t.

Let me rewrite the equations:

Equation 1: 9 = [s*(240 - t)] / 60 → Multiply both sides by 60: 540 = s*(240 - t)

Equation 2: 9 = [(s + 2)*(144 - t)] / 60 → Multiply both sides by 60: 540 = (s + 2)*(144 - t)

So now we have:

1) 540 = s*(240 - t)

2) 540 = (s + 2)*(144 - t)

Therefore, set the two expressions equal to each other:

s*(240 - t) = (s + 2)*(144 - t)

Let me expand the right-hand side:

(s + 2)*(144 - t) = s*(144 - t) + 2*(144 - t) = 144s - s t + 288 - 2t

So the equation becomes:

s*(240 - t) = 144s - s t + 288 - 2t

Let me bring all terms to the left side:

s*(240 - t) - [144s - s t + 288 - 2t] = 0

Compute term by term:

First term: s*(240 - t) = 240s - s t

Subtract the other terms:

-144s + s t - 288 + 2t = 0

Wait, let me do it step by step:

Left side after subtraction:

240s - s t - 144s + s t - 288 + 2t = 0

Wait, let me check:

Original left side after expanding:

240s - s t

Minus the terms in the bracket:

-144s + s t - 288 + 2t

Wait, actually, the entire expression is:

[240s - s t] - [144s - s t + 288 - 2t] = 0

So distribute the negative sign:

240s - s t -144s + s t -288 + 2t = 0

Now, let's combine like terms:

240s -144s = 96s

-s t + s t = 0

Then -288 + 2t

So overall:

96s - 288 + 2t = 0

So 96s + 2t = 288

We can divide both sides by 2:

48s + t = 144

So that's equation (3): 48s + t = 144

Now, from equation 1: 540 = s*(240 - t)

We can perhaps express t from equation 3 and substitute into equation 1.

From equation 3: t = 144 - 48s

Plug into equation 1:

540 = s*(240 - (144 - 48s)) = s*(240 -144 +48s) = s*(96 +48s)

So:

540 = 48s^2 + 96s

Bring all terms to one side:

48s^2 + 96s - 540 = 0

Divide all terms by 12 to simplify:

4s^2 + 8s - 45 = 0

Hmm, quadratic equation: 4s² +8s -45 =0

Let me solve for s using quadratic formula.

s = [-8 ± sqrt(64 + 4*4*45)]/(2*4)

Compute discriminant:

64 + 4*4*45 = 64 + 16*45 = 64 + 720 = 784

sqrt(784)=28

Thus,

s = [ -8 ±28 ] /8

We can discard the negative solution because speed can't be negative.

So,

s = (-8 +28)/8 = 20/8 = 2.5 km/h

Wait, 20 divided by 8 is 2.5? Wait 20/8 is 2.5? 8*2=16, 20-16=4, so 2 and 4/8, which is 2.5. Yes.

So s = 2.5 km/h

Then, from equation 3: 48s + t =144

So t =144 -48*(2.5)

Compute 48*2.5: 48*2=96, 48*0.5=24, so total 120.

Thus, t =144 -120 =24 minutes.

So t is 24 minutes.

Let me check if this makes sense.

First scenario: speed s=2.5 km/h, total time 4 hours (240 minutes). The coffee shop time is 24 minutes, so walking time is 240 -24 =216 minutes, which is 3.6 hours.

Distance = speed * time = 2.5 km/h * 3.6 h = 9 km. Correct.

Second scenario: speed s+2 =4.5 km/h. Total time 2h24m =144 minutes. Coffee time still 24 minutes, so walking time is 144-24=120 minutes =2 hours.

Distance =4.5 km/h *2 h =9 km. Correct.

Okay, so that checks out.

Now, the question is, when she walks at s +1/2 km/h. Since s is 2.5, then s +0.5 is 3 km/h.

So new speed is 3 km/h. Need to find total time, including t=24 minutes.

First, compute the walking time. Distance is 9 km, speed is 3 km/h, so time is 9/3 =3 hours. That's 180 minutes. Then add the coffee time of 24 minutes, so total time is 180 +24=204 minutes.

Wait, but let me confirm again.

Wait, the total time is walking time plus coffee time. The walking time at 3 km/h is 3 hours, which is 180 minutes. Adding 24 minutes gives 204 minutes total. So 204 minutes is the answer?

Wait, but let me check again with the equations to be sure.

Alternatively, let me use the same approach as before.

Let me denote the new speed as s + 1/2 = 2.5 +0.5 =3 km/h.

Let the total time be T minutes. Then the walking time is (T - t) minutes, which is (T -24)/60 hours. The distance is 9 km, so:

9 = 3 * [(T -24)/60]

Solve for T:

Multiply both sides by 60: 540 =3*(T -24)

Divide both sides by 3: 180 = T -24

Thus, T =180 +24 =204 minutes. So that's correct.

Therefore, the answer is 204 minutes. So boxed answer is \boxed{204}

Wait, but let me just think again if there's another way this could be interpreted. The problem says "the walk takes her...", so the total time including the coffee shop. So yes, that's exactly what we calculated. So I think that's correct.

Alternatively, maybe I made an error in units somewhere? Let me check again.

First scenario:

s =2.5 km/h, total time 4 hours (240 minutes). Coffee time 24 minutes, so walking time 216 minutes (3.6 hours). 2.5 * 3.6 =9. Correct.

Second scenario:

s+2 =4.5 km/h, total time 144 minutes (2.4 hours). Coffee 24 minutes, so walking time 120 minutes (2 hours). 4.5 *2 =9. Correct.

Third scenario:

s +0.5 =3 km/h. Walking time is 9/3=3 hours=180 minutes. Total time 180 +24=204 minutes. So yes, that's right.

Therefore, the answer is 204. Seems straightforward once the equations are set up. I think that's correct.

**Final Answer**
\boxed{204}
</think>

Aya walks 9 kilometers each morning and stops at a coffee shop for \( t \) minutes. We need to determine the total time taken when she walks at \( s + \frac{1}{2} \) kilometers per hour, including the coffee shop stop.

First, we set up the equations based on the given information:

1. When walking at \( s \) km/h, the total time is 4 hours (240 minutes):
   \[
   9 = s \times \frac{240 - t}{60}
   \]
   Simplifying, we get:
   \[
   540 = s \times (240 - t)
   \]

2. When walking at \( s + 2 \) km/h, the total time is 2 hours 24 minutes (144 minutes):
   \[
   9 = (s + 2) \times \frac{144 - t}{60}
   \]
   Simplifying, we get:
   \[
   540 = (s + 2) \times (144 - t)
   \]

Equating the two expressions for 540:
\[
s \times (240 - t) = (s + 2) \times (144 - t)
\]

Expanding and simplifying:
\[
240s - st = 144s - st + 288 - 2t
\]
\[
96s = 288 - 2t
\]
\[
48s + t = 144
\]

Solving for \( t \) from the equation \( 48s + t = 144 \):
\[
t = 144 - 48s
\]

Substituting \( t \) back into the first equation:
\[
540 = s \times (240 - (144 - 48s))
\]
\[
540 = s \times (96 + 48s)
\]
\[
48s^2 + 96s - 540 = 0
\]

Solving the quadratic equation:
\[
s = \frac{-8 + \sqrt{784}}{8} = \frac{20}{8} = 2.5 \text{ km/h}
\]

Finding \( t \):
\[
t = 144 - 48 \times 2.5 = 24 \text{ minutes}
\]

For the speed \( s + \frac{1}{2} = 3 \) km/h:
\[
\text{Walking time} = \frac{9}{3} = 3 \text{ hours} = 180 \text{ minutes}
\]
\[
\text{Total time} = 180 + 24 = 204 \text{ minutes}
\]

Thus, the number of minutes the walk takes her, including the coffee shop stop, is \(\boxed{204}\).


Solution:
From the tangency condition we have $\let\angle BCD = \let\angle CBD = \let\angle A$. With LoC we have $\cos(A) = \frac{25+100-81}{2*5*10} = \frac{11}{25}$ and $\cos(B) = \frac{81+25-100}{2*9*5} = \frac{1}{15}$. Then, $CD = \frac{\frac{9}{2}}{\cos(A)} = \frac{225}{22}$. Using LoC we can find $AD$: $AD^2 = AC^2 + CD^2 - 2(AC)(CD)\cos(A+C) = 10^2+(\frac{225}{22})^2 + 2(10)\frac{225}{22}\cos(B) = 100 + \frac{225^2}{22^2} + 2(10)\frac{225}{22}*\frac{1}{15} = \frac{5^4*13^2}{484}$. Thus, $AD = \frac{5^2*13}{22}$. By Power of a Point, $DP*AD = CD^2$ so $DP*\frac{5^2*13}{22} = (\frac{225}{22})^2$ which gives $DP = \frac{5^2*9^2}{13*22}$. Finally, we have $AP = AD - DP = \frac{5^2*13}{22} - \frac{5^2*9^2}{13*22} = \frac{100}{13} \rightarrow \boxed{113}$.
~angie.
We know $AP$ is the symmedian, which implies $\triangle{ABP}\sim \triangle{AMC}$ where $M$ is the midpoint of $BC$. By Appolonius theorem, $AM=\frac{13}{2}$. Thus, we have $\frac{AP}{AC}=\frac{AB}{AM}, AP=\frac{100}{13}\implies \boxed{113}$
~Bluesoul
Extend sides $\overline{AB}$ and $\overline{AC}$ to points $E$ and $F$, respectively, such that $B$ and $C$ are the feet of the altitudes in $\triangle AEF$. Denote the feet of the altitude from $A$ to $\overline{EF}$ as $X$, and let $H$ denote the orthocenter of $\triangle AEF$. Call $M$ the midpoint of segment $\overline{EF}$. By the Three Tangents Lemma, we have that $MB$ and $MC$ are both tangents to $(ABC)$ $\implies$ $M = D$, and since $M$ is the midpoint of $\overline{EF}$, $MF = MB$. Additionally, by angle chasing, we get that:
\[\angle ABC \cong \angle AHC \cong \angle EHX\]
Also,
\[\angle EHX = 90 ^\circ - \angle HEF = 90 ^\circ - (90 ^\circ - \angle AFE) = \angle AFE\]
Furthermore,
\[AB = AF \cdot \cos(A)\]
From this, we see that $\triangle ABC \sim \triangle AFE$ with a scale factor of $\cos(A)$. By the Law of Cosines,
\[\cos(A) = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}\]
Thus, we can find that the side lengths of $\triangle AEF$ are $\frac{250}{11}, \frac{125}{11}, \frac{225}{11}$. Then, by Stewart's theorem, $AM = \frac{13 \cdot 25}{22}$. By Power of a Point,
\[\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MP}\]
\[\frac{225}{22} \cdot \frac{225}{22} = \overline{MP} \cdot \frac{13 \cdot 25}{22} \implies \overline{MP} = \frac{225 \cdot 9}{22 \cdot 13}\]
Thus,
\[AP = AM - MP = \frac{13 \cdot 25}{22} - \frac{225 \cdot 9}{22 \cdot 13} = \frac{100}{13}\]
Therefore, the answer is $\boxed{113}$.
~mathwiz_1207
Connect lines $\overline{PB}$ and $\overline{PC}$. From the angle by tanget formula, we have $\angle PBD = \angle DAB$. Therefore by AA similarity, $\triangle PBD \sim \triangle BAD$. Let $\overline{BP} = x$. Using ratios, we have \[\frac{x}{5}=\frac{BD}{AD}.\] Similarly, using angle by tangent, we have $\angle PCD = \angle DAC$, and by AA similarity, $\triangle CPD \sim \triangle ACD$. By ratios, we have \[\frac{PC}{10}=\frac{CD}{AD}.\] However, because $\overline{BD}=\overline{CD}$, we have \[\frac{x}{5}=\frac{PC}{10},\] so $\overline{PC}=2x.$ Now using Law of Cosines on $\angle BAC$ in triangle $\triangle ABC$, we have \[9^2=5^2+10^2-100\cos(\angle BAC).\] Solving, we find $\cos(\angle BAC)=\frac{11}{25}$. Now we can solve for $x$. Using Law of Cosines on $\triangle BPC,$ we have
\begin{align*}
81&=x^2+4x^2-4x^2\cos(180-\angle BAC) \\
&= 5x^2+4x^2\cos(BAC). \\
\end{align*}
Solving, we get $x=\frac{45}{13}.$ Now we have a system of equations using Law of Cosines on $\triangle BPA$ and $\triangle CPA$, \[AP^2=5^2+\left(\frac{45}{13}\right)^2 -(10) \left(\frac{45}{13} \right)\cos(ABP)\]
\[AP^2=10^2+4 \left(\frac{45}{13} \right)^2 + (40) \left(\frac{45}{13} \right)\cos(ABP).\]
Solving, we find $\overline{AP}=\frac{100}{13}$, so our desired answer is $100+13=\boxed{113}$.
~evanhliu2009
Following from the law of cosines, we can easily get $\cos A = \frac{11}{25}$, $\cos B = \frac{1}{15}$, $\cos C = \frac{13}{15}$.
Hence, $\sin A = \frac{6 \sqrt{14}}{25}$, $\cos 2C = \frac{113}{225}$, $\sin 2C = \frac{52 \sqrt{14}}{225}$.
Thus, $\cos \left( A + 2C \right) = - \frac{5}{9}$.
Denote by $R$ the circumradius of $\triangle ABC$.
In $\triangle ABC$, following from the law of sines, we have $R = \frac{BC}{2 \sin A} = \frac{75}{4 \sqrt{14}}$.
Because $BD$ and $CD$ are tangents to the circumcircle $ABC$, $\triangle OBD \cong \triangle OCD$ and $\angle OBD = 90^\circ$.
Thus, $OD = \frac{OB}{\cos \angle BOD} = \frac{R}{\cos A}$.
In $\triangle AOD$, we have $OA = R$ and $\angle AOD = \angle BOD + \angle AOB = A + 2C$.
Thus, following from the law of cosines, we have
\begin{align*}
AD & = \sqrt{OA^2 + OD^2 - 2 OA \cdot OD \cos \angle AOD} \\
& = \frac{26 \sqrt{14}}{33} R.
\end{align*}

Following from the law of cosines,
\begin{align*}
\cos \angle OAD & = \frac{AD^2 + OA^2 - OD^2}{2 AD \cdot OA} \\
& = \frac{8 \sqrt{14}}{39} .
\end{align*}

Therefore,
\begin{align*}
AP & = 2 OA \cos \angle OAD \\
& = \frac{100}{13} .
\end{align*}

Therefore, the answer is $100 + 13 = \boxed{\textbf{(113) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Answer: 113

<｜begin▁of▁sentence｜><｜User｜>Let $ABC$ be a triangle inscribed in circle $\omega$. Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$, and let $\overline{AD}$ intersect $\omega$ at $P$. If $AB=5$, $BC=9$, and $AC=10$, $AP$ can be written as the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers. Find $m + n$.
[asy] import olympiad;  unitsize(15);  pair A, B, C, D, E, F, P, O;  C = origin; A = (10,0); B = (7.8, 4.4899); draw(A--B--C--cycle); draw(A..B..C..cycle, red+dotted);  O = circumcenter(A, B, C);  E = rotate(90,B) * (O); F = rotate(90,C) * (O);  D = IP(B..E + (B-E)*4, C..F + (C-F)*-3);  draw(B--D--C--D--A);  P = IP(D..A, A..B..C);  dot(A); dot(B); dot(C); dot(D); dot(P); label("$A$", A, dir(335)); label("$B$", B, dir(65)); label("$C$", C, dir(200)); label("$D$", D, dir(135)); label("$P$", P, dir(235)); [/asy]<｜Assistant｜><think>

Okay, so I have this geometry problem here. Let me try to parse it step by step. The problem is about triangle ABC inscribed in a circle ω. The tangents to ω at points B and C intersect at point D. Then, the line AD intersects the circle ω again at point P. We are given the side lengths AB = 5, BC = 9, and AC = 10. The goal is to find the length AP expressed as a fraction m/n where m and n are coprime, and then compute m + n.

Hmm, first, I need to recall some geometry concepts here. Since D is the intersection of the tangents at B and C to the circumcircle ω of triangle ABC, D is the exsimilicenter or the symmedian point? Wait, actually, the intersection of two tangents to a circle from a point is related to the concept of the symmedian. Alternatively, D is the ex-pole of line BC or something? Maybe I need to recall that the tangents from a point to a circle are equal in length. So DB = DC because they are both tangents from D to ω. Wait, no, actually, the two tangent segments from a single point to a circle are equal. Since D is the intersection of the tangents at B and C, then DB and DC are tangent to the circle, so DB = DC. So D is equidistant from B and C. That might be useful.

Alternatively, perhaps using power of a point. The power of point D with respect to ω is DB² = DC², which is equal to the power of D, so that's consistent. Also, since AD intersects the circle again at P, then AP * AD = power of A with respect to ω? Wait, no, actually, the power of point A with respect to ω is zero because A is on the circle. Hmm, maybe I need to use some other theorem here.

Alternatively, maybe using harmonic division, or projective geometry concepts. Alternatively, since D is the intersection of the two tangents, perhaps triangle DBC is related to the triangle ABC in some way. Maybe some similar triangles?

Alternatively, perhaps using coordinates or trigonometry. Since the sides are given, maybe we can compute coordinates of the points and then find AP?

Alternatively, using inversion. Hmm, inversion might be overkill, but perhaps possible.

Alternatively, using Ceva's theorem or Menelaus' theorem?

Alternatively, since D is the exsymcon or something? Wait, maybe the polar of D passes through some point?

Alternatively, perhaps using the concept of symmedian. Since the tangent at B is the symmedian of triangle DBC? Wait, maybe not. Let me think again.

Alternatively, since D is the intersection of the tangents at B and C, then line AD is the symmedian of triangle ABC? Wait, maybe not exactly, but perhaps there is a relation here.

Alternatively, perhaps using the power of point D with respect to the circle. Since D lies outside the circle, the power of D is DB² = DC² = DP * DA. Wait, yes! Because AD passes through the circle at A and P, so the power of D with respect to ω is equal to DB² = DC² = DP * DA. That seems important.

So, Power of a Point theorem says that for point D outside the circle, the power is equal to DB² (since DB is tangent) and also equal to DP * DA, where DP and DA are the lengths along the secant line AD. Therefore, DB² = DP * DA. So if I can compute DB and express DP in terms of AP, maybe I can set up an equation.

Let me note that AP is the segment from A to P, and since P is on the circle, AP is a chord. The length AD is AP + PD, but actually, since D is outside the circle, AD is from A to D, passing through P, so AP is from A to P, and then PD is from P to D. Therefore, AD = AP + PD. But in terms of signed lengths, depending on direction, but since all lengths are positive, we can write AD = AP + PD. Therefore, PD = AD - AP. Therefore, substituting into the power of point D: DB² = DP * DA = (AD - AP) * AD. Hmm, but that might not be directly helpful unless we can relate AP and AD.

Alternatively, maybe express AP in terms of AD. Let me denote AP = x, so PD = AD - x. Then, according to the power of point D: DB² = x*(AD). Wait, no, wait. Wait, the power of D is equal to DP * DA. Wait, actually, the formula is: if a line through D intersects the circle at two points, say P and A (with P closer to D?), then DP * DA = power of D. Wait, actually, the correct formula is that if the secant from D passes through the circle at two points, say P and A, with P closer to D, then DP * DA = power. Wait, but in our case, the line AD starts at A (which is on the circle) and goes to D (outside), so the two intersection points with the circle are A and P. Therefore, the power of D is equal to DP * DA, where DP is the distance from D to P, and DA is the distance from D to A. Wait, actually, the formula is: power of D = DP * DA, where DP is the entire length from D to A, but since A is on the circle, the power is DP * DA = (DP) * (DA). Wait, perhaps I need to clarify.

Wait, let me recall the exact statement. The power of a point D with respect to a circle is equal to the product of the lengths from D to the two intersection points of any line through D with the circle. So in this case, the line AD passes through A (which is on the circle) and P (the other intersection point). Therefore, the power of D is equal to DP * DA. Wait, but DA is the distance from D to A, and DP is the distance from D to P. Since A is on the circle, then DA is the length from D to A, and DP is from D to P. Since P is between A and D? Wait, actually, since D is outside the circle, and AD is a line from A (on the circle) to D (outside), then the two intersection points of line AD with the circle are A and P, so starting from D, going towards A, you first meet P, then A. Therefore, DP is the distance from D to P, and PA is from P to A. Therefore, DA = DP + PA. Therefore, the power of D is DP * DA = DP*(DP + PA). But since A is on the circle, the power of D is also equal to DB². Therefore, DB² = DP*(DP + PA). Hmm, so if I let PA = x, then DP = DA - x. Wait, but DA is the distance from D to A, so DA = DP + PA, so DP = DA - PA. Therefore, substituting, DB² = (DA - PA) * DA. Therefore, DB² = DA² - PA * DA. Hmm, but this seems a bit convoluted. Maybe I need another equation involving PA and DA.

Alternatively, perhaps express PA in terms of DA. Let me denote PA = x, so DA = x + DP, but I don't know DP yet. Alternatively, perhaps we can find DA in terms of other quantities?

Alternatively, maybe use Stewart's theorem on triangle ABD or something? Wait, but I need more information.

Alternatively, perhaps use coordinates. Let me try to set up coordinate system.

Let me consider placing triangle ABC in coordinate plane. Let me see, given the sides AB = 5, BC = 9, AC = 10. Let me note that the sides are 5, 9, 10. Let me check if this is a right triangle? Let me see: 5² + 9² = 25 + 81 = 106, which is not 10² = 100, so no. 5² + 10² = 125 vs 9² = 81, no. So it's a scalene triangle.

Perhaps place point A at (0,0), point C at (10,0), since AC = 10. Then point B is somewhere in the plane. Let me compute coordinates of B.

Let me denote coordinates: Let’s set point A at (0,0), point C at (10,0). Let point B have coordinates (x,y). Then, AB = 5, so distance from A(0,0) to B(x,y) is sqrt(x² + y²) = 5, so x² + y² = 25. The distance from B to C(10,0) is 9, so sqrt((x -10)^2 + y²) = 9, so (x -10)^2 + y² = 81. Also, AC is 10, which is already set.

So we have two equations:

1. x² + y² = 25,

2. (x -10)^2 + y² = 81.

Subtracting first equation from the second:

(x -10)^2 + y² - x² - y² = 81 -25 => (x² -20x +100) -x² = 56 => -20x +100 =56 => -20x = -44 => x = 44/20 = 11/5 = 2.2.

Then, substituting back x = 11/5 into first equation:

(11/5)^2 + y² =25 => 121/25 + y² = 25 => y² = 25 - 121/25 = (625 -121)/25 = 504/25 => y = sqrt(504)/5 = (sqrt(504))/5. Simplify sqrt(504): 504 = 16*31.5? Wait, 504 divided by 16 is 31.5, which is not integer. Wait, 504 = 16*31.5? Wait, 504 divided by 9 is 56, so 504 = 9*56 = 9*7*8 = 9*7*8. So sqrt(504) = sqrt(9*56) = 3*sqrt(56) = 3*sqrt(4*14) = 3*2*sqrt(14) = 6*sqrt(14). Therefore, sqrt(504)/5 = (6√14)/5. Therefore, coordinates of B are (11/5, (6√14)/5). Hmm, okay. So coordinates are:

A: (0,0)

C: (10,0)

B: (11/5, (6√14)/5 )

Now, need to find the circumcircle ω of triangle ABC. Let me compute its equation.

The circumcircle can be found by finding the perpendicular bisectors of two sides and finding their intersection.

Let me compute the perpendicular bisector of AB and AC, or maybe AB and BC. Let me see.

First, let me compute the midpoint of AB: midpoint M1 is ((0 + 11/5)/2, (0 + (6√14)/5)/2 ) = (11/10, (3√14)/5 )

The slope of AB is ( (6√14)/5 - 0 ) / (11/5 - 0 ) = (6√14)/5 divided by 11/5 = 6√14 /11. Therefore, the perpendicular bisector will have slope -11/(6√14).

Similarly, the perpendicular bisector of AC: midpoint of AC is (5,0). The slope of AC is 0 (since it's along the x-axis), so the perpendicular bisector is vertical line x =5.

Therefore, the perpendicular bisector of AC is x=5. So the center of the circle lies on x=5. Now, let's find the perpendicular bisector of AB and find its intersection with x=5.

The perpendicular bisector of AB passes through M1 (11/10, 3√14/5 ) and has slope -11/(6√14). Let me write its equation.

The equation is: y - 3√14/5 = (-11)/(6√14) (x - 11/10 )

We need to find where this line intersects x =5. Let me substitute x=5 into this equation.

So, y = 3√14/5 + (-11)/(6√14)*(5 - 11/10 )

Compute 5 - 11/10 = 50/10 -11/10 = 39/10.

Therefore,

y = 3√14/5 + (-11)/(6√14)*(39/10 )

Simplify the second term:

(-11 * 39)/(6√14 *10 ) = (-429)/(60√14 ) = (-143)/(20√14 )

Therefore,

y = (3√14)/5 - (143)/(20√14 )

To combine these terms, let me rationalize the denominators.

First term: 3√14 /5 = (3√14 * √14)/ (5√14) = (3*14)/ (5√14) = 42/(5√14)

Wait, but maybe better to get a common denominator. Let me see:

Let me write both terms over 20√14:

First term: (3√14)/5 = (3√14 * 4√14)/ (5 * 4√14) = Wait, perhaps another approach.

Alternatively, let me compute numerically, but maybe it's better to keep it symbolic.

Alternatively, let me compute the y-coordinate:

Let me write:

y = (3√14)/5 - (143)/(20√14 )

Multiply numerator and denominator of the second term by √14 to rationalize:

= (3√14)/5 - (143√14)/(20*14 )

= (3√14)/5 - (143√14)/280

Convert to common denominator, which is 280:

First term: (3√14)/5 = (3√14 * 56)/280 = 168√14 /280

Second term: 143√14 /280

Therefore,

y = (168√14 -143√14)/280 = (25√14)/280 = (5√14)/56

Therefore, the center of the circle is at (5, 5√14 /56 )

Wait, let me check the calculation again because that seems a bit messy. Let me verify step by step.

First, the perpendicular bisector of AB:

Midpoint M1 is (11/10, 3√14/5 )

Slope of perpendicular bisector is -11/(6√14 )

Equation: y - 3√14/5 = (-11)/(6√14)(x - 11/10 )

At x =5,

y = 3√14/5 + (-11)/(6√14)*(5 - 11/10 )

Compute 5 - 11/10 = 50/10 -11/10 = 39/10.

So,

y = 3√14/5 - (11 * 39)/(6√14 *10 )

Compute numerator of the second term: 11*39 = 429

Denominator: 6√14 *10 = 60√14

Thus,

Second term is -429/(60√14 )

So,

y = 3√14/5 - 429/(60√14 )

Let me express both terms with denominator 60√14:

First term: 3√14 /5 = (3√14 * 12√14)/ (5 *12√14) Hmm, perhaps better to convert 3√14/5 to over 60√14:

Multiply numerator and denominator by 12:

(3√14/5) = (3√14 *12)/(5*12) = (36√14)/60. But that's over 60, not 60√14. Hmm, maybe another approach.

Alternatively, let me rationalize the second term:

-429/(60√14 ) = -429√14/(60*14) = -429√14 /840

Similarly, the first term is 3√14 /5 = (3√14 * 168)/840 = (504√14)/840

Therefore,

y = (504√14 - 429√14)/840 = (75√14)/840 = (15√14)/168 = (5√14)/56

Yes, so that gives y = 5√14 /56. So the center O is at (5, 5√14 /56 )

Hmm, okay. That seems correct. Let me note that.

Therefore, the center of the circumcircle ω is at (5, 5√14 /56 ). Now, we can compute the radius of the circle. Let me compute the distance from center O to point A (0,0):

OA² = (5 -0)^2 + (5√14 /56 -0)^2 = 25 + (25*14)/(56² )

Compute 25*14 = 350, and 56² = 3136.

Thus,

OA² =25 + 350/3136 = 25 + (350 ÷ 14)/(3136 ÷14) = 25 + 25/224 ≈ but maybe keep as fractions.

25 is 25/1, so common denominator 224:

25 = 25*224/224 = 5600/224,

Thus OA² = (5600 + 25)/224 = 5625/224. Therefore, OA = sqrt(5625/224) = 75/sqrt(224). Hmm, sqrt(224) = sqrt(16*14) = 4√14, so OA = 75/(4√14) = (75√14)/(4*14) = (75√14)/56. So radius is 75√14 /56.

Alternatively, maybe I need the equation of the circle. Let me write it as (x -5)^2 + (y - 5√14/56)^2 = (75√14 /56)^2.

Now, the next step is to find point D, which is the intersection of the tangents at B and C to the circle ω.

To find point D, we can find the equations of the two tangents at B and C, then find their intersection.

First, let me find the equation of the tangent at point B.

The tangent at a point on a circle is perpendicular to the radius at that point. So the tangent at B is perpendicular to the radius OB, where O is the center (5, 5√14/56).

Coordinates of B are (11/5, (6√14)/5 ). Let me compute the vector from O to B:

O is (5, 5√14/56 ), so coordinates difference:

x-coordinate: 11/5 -5 = 11/5 -25/5 = -14/5,

y-coordinate: (6√14)/5 - 5√14/56 = (6√14 *56 -5√14 *5)/ (5*56 )

Compute numerator:

6√14 *56 = 336√14,

5√14 *5 =25√14,

Thus numerator: 336√14 -25√14 = 311√14,

Denominator: 280,

Therefore, the y-component is 311√14 /280.

Therefore, the direction vector from O to B is (-14/5, 311√14 /280 ). The tangent at B is perpendicular to this vector.

Therefore, the slope of OB is (Δy)/(Δx) = (311√14 /280 ) / (-14/5 ) = (311√14 /280 ) * (-5/14 ) = (-311√14 *5 )/(280*14 )

Simplify denominators: 280*14 = 3920,

Numerator: -1555√14,

Thus slope of OB is (-1555√14)/3920. Let me see if this can be simplified. Hmm, perhaps not easily. Alternatively, maybe compute the slope of the tangent as the negative reciprocal.

The slope of the tangent at B is perpendicular to OB, so its slope is the negative reciprocal of the slope of OB.

Let me denote m_OB = (Δy)/(Δx) = [ (6√14/5 -5√14/56 ) ] / (11/5 -5 )

Wait, perhaps it's better to compute numerically.

Alternatively, perhaps there is a better way. Let me recall that the equation of the tangent to a circle at a point (x1,y1) is (x1 - a)(x - a) + (y1 - b)(y - b) = r²? Wait, no, the standard equation for a circle with center (h,k) is (x - h)^2 + (y - k)^2 = r^2. The tangent at point (x1,y1) on the circle is (x1 - h)(x - h) + (y1 - k)(y - k) = r^2? Wait, no, actually, the equation is (x1 - h)(x - h) + (y1 - k)(y - k) = r². Wait, but actually, that's the equation for the tangent line at (x1,y1). Wait, let me confirm.

Wait, actually, the equation of the tangent line at point (x1,y1) on the circle (x - h)^2 + (y - k)^2 = r^2 is:

(x1 - h)(x - h) + (y1 - k)(y - k) = r^2.

Wait, but actually, no, that's the equation for the tangent line. Wait, let me check:

Wait, the standard formula is:

For a circle with center (h,k), the tangent at point (x1,y1) is:

(x1 - h)(x - h) + (y1 - k)(y - k) = r^2.

Wait, but actually, that's the equation of the circle itself. Wait, no, perhaps not. Wait, let me think of the tangent line equation.

Alternatively, the tangent line can be written as:

(x1 - h)(x - h) + (y1 - k)(y - k) = r^2 ?

Wait, no, that would be the same as the circle equation. Wait, perhaps I need to use the point-form of the tangent.

Alternatively, the tangent line at (x1,y1) is:

(x1 - h)(x - h) + (y1 - k)(y - k) = r^2 ?

Wait, but substituting (x1,y1) into that equation gives (x1 - h)^2 + (y1 - k)^2 = r^2, which is true, but that's the equation of the circle. So that can't be the tangent line. Hmm, perhaps I made a mistake.

Wait, actually, the correct equation for the tangent line at (x1,y1) on the circle (x - h)^2 + (y - k)^2 = r^2 is:

(x1 - h)(x - h) + (y1 - k)(y - k) = r^2.

Wait, but that is the same as the circle equation, so that can't be. Wait, no, actually, no. Wait, the tangent line equation is different. Let me recall that the tangent line is perpendicular to the radius. So, the line through (x1,y1) with slope perpendicular to the radius.

Alternatively, the equation can be written as:

(x1 - h)(x - h) + (y1 - k)(y - k) = r^2 ?

Wait, perhaps not. Let me check with a simple example. Let the circle be x² + y² = r², so center at (0,0). The tangent at (x1,y1) is x x1 + y y1 = r². Yes, that's correct. So in general, for circle (x - h)^2 + (y - k)^2 = r^2, the tangent at (x1,y1) is (x1 - h)(x - h) + (y1 - k)(y - k) = r^2. Yes, that's correct. So my initial thought was right.

Therefore, the tangent at point B (x1,y1) is:

(x_B - h)(x - h) + (y_B - k)(y - k) = r^2,

where (h,k) is the center (5,5√14/56 ), and (x_B,y_B) is point B (11/5, 6√14/5 ).

Therefore, plugging in:

( (11/5 -5)(x -5) + ( (6√14/5 -5√14/56 )(y -5√14/56 ) ) = r^2.

Wait, let me compute each term step by step.

First, compute (x_B - h):

x_B - h = 11/5 -5 = 11/5 -25/5 = -14/5.

Similarly, (y_B - k):

y_B - k = (6√14/5 ) - (5√14/56 ) = let's compute this:

Common denominator is 5*56 = 280.

So,

6√14/5 = (6√14 *56)/280 = 336√14 /280,

5√14/56 = (5√14 *5)/280 =25√14 /280,

Thus,

y_B -k = 336√14/280 -25√14/280 = 311√14 /280.

Therefore, the equation of the tangent at B is:

(-14/5)(x -5) + (311√14 /280)(y -5√14/56 ) = r^2.

Similarly, the tangent at point C (10,0):

First, coordinates of C are (10,0). Let me compute the tangent at C.

Similarly, using the same formula:

(x_C - h)(x - h) + (y_C - k)(y - k) = r^2.

Compute (x_C - h) =10 -5 =5,

(y_C -k ) =0 -5√14/56 = -5√14/56.

Thus, the tangent at C is:

5*(x -5) + (-5√14/56)(y -5√14/56 ) = r^2.

Therefore, now we have equations of both tangents at B and C. We need to find their intersection point D.

This seems quite involved with all these fractions. Maybe I need to compute r² first.

Earlier, we had OA² = (75√14 /56 )², so r² = OA² = (75√14 /56 )² = (75² *14)/ (56² )

Compute 75² = 5625,

Thus, r² = (5625 *14)/ (3136 ) = 78750 / 3136.

Hmm, okay. Let me note that.

So, the equation of tangent at B is:

(-14/5)(x -5) + (311√14 /280)(y -5√14/56 ) = 78750/3136.

Similarly, tangent at C is:

5(x -5) + (-5√14/56)(y -5√14/56 ) = 78750/3136.

These equations look really messy. Maybe there's a smarter way?

Alternatively, since D is the exsymcon or the symmedian point? Wait, perhaps using pole-polar concepts.

Alternatively, since D is the intersection of the two tangents, then line DC and DB are tangent to the circle, so DB = DC as lengths. Wait, but coordinates might still be necessary.

Alternatively, perhaps use parametric equations for the tangent lines and solve for D.

Alternatively, maybe use vector approaches?

Alternatively, perhaps use the fact that D lies on the polar of the Gergonne point or something? Hmm, perhaps not.

Alternatively, since this is getting too algebraic, maybe there is a synthetic approach.

Wait, let me think again. The problem gives sides AB=5, BC=9, AC=10. We need to find AP where P is the second intersection of AD with the circumcircle, and D is the intersection of the tangents at B and C.

Perhaps using power of a point A with respect to some circle? Wait, but A is on the circumcircle, so its power is zero.

Alternatively, use inversion with respect to the circumcircle. Inverting the figure might simplify things, but I might need to think more.

Alternatively, use harmonic division. Since D is the intersection of tangents at B and C, then perhaps BD and CD are tangents, so BD=CD, and maybe some harmonic bundle?

Alternatively, use Ceva's theorem in some triangle. Let me see.

Alternatively, use the concept that AD is the symmedian of triangle ABC. Wait, because the tangents at B and C meet at D, so D is the symmedian point? Wait, the symmedian point is the intersection of the symmedians, which are the reflections of the medians over the angle bisectors. Alternatively, the symmedian can be constructed as the tangent at A to the circumcircle, but not sure.

Alternatively, in triangle ABC, the point D is the exsymmedian point? Wait, perhaps the symmedian point related to BC.

Alternatively, perhaps AD is the symmedian from A. If that is the case, then the ratio of AB²/AC² = ratio of segments on BC. Wait, but not sure.

Alternatively, the symmedian reflection property: if AD is the symmedian, then (BD/DC) = (AB²/AC²). Wait, but BD and DC here are the lengths from D to B and C, but D is outside the triangle.

Alternatively, since D is the intersection of the tangents at B and C, then AD is the symmedian of triangle ABC. Let me check this.

Wait, in triangle ABC, the symmedian from A is the reflection of the median over the angle bisector. Alternatively, the symmedian can be defined as the locus of points whose distances to AB and AC are proportional to AB and AC.

Alternatively, the symmedian can be constructed by taking a point D such that DB/DC = AB²/AC². Wait, that might be the case here.

Wait, since D is the intersection of the tangents at B and C, then by the property of symmedian, the point D lies on the symmedian of angle A if and only if DB/DC = AB²/AC². Let me check that.

Wait, according to the symmedian properties, if D is on the symmedian, then DB/DC = AB²/AC². Let me see if that holds here.

Given that D is the intersection of the tangents at B and C, then by the power of point D, DB = DC (since both are tangent lengths from D to the circle). Wait, but DB = DC, so DB/DC =1. However, AB²/AC² =25/100 =1/4. So unless 1 =1/4, which is not, so that can't be. Therefore, AD is not the symmedian. Hmm, so that approach might not work.

Alternatively, perhaps use trigonometric Ceva's theorem. Since D is the intersection of the tangents, perhaps some angle relations?

Alternatively, let me think of using coordinates again, even though it's tedious. Let me try to write equations for the two tangent lines and solve for D.

First, let me rewrite the tangent at B equation:

Equation of tangent at B:

(-14/5)(x -5) + (311√14 /280)(y -5√14/56 ) = 78750/3136.

Similarly, tangent at C:

5(x -5) + (-5√14/56)(y -5√14/56 ) = 78750/3136.

Let me first work on the tangent at C equation, as it might be simpler.

Let me denote equation of tangent at C:

5(x -5) + (-5√14/56)(y -5√14/56 ) = 78750/3136.

Let me expand this:

5x -25 + (-5√14/56)y + (5√14/56)(5√14/56 ) = 78750/3136.

Compute the term (5√14/56)(5√14/56 ):

= (25 *14)/ (56² ) = 350 / 3136.

Therefore, the equation becomes:

5x -25 - (5√14/56)y + 350/3136 = 78750/3136.

Multiply all terms by 3136 to eliminate denominators:

5x*3136 -25*3136 - (5√14/56)y*3136 +350 =78750.

Compute each term:

5x*3136 = 15680x,

-25*3136 = -78400,

-(5√14/56)y *3136 = -5√14 * (3136/56 ) y = -5√14 *56 y = -280√14 y,

+350,

= 78750.

So putting together:

15680x -78400 -280√14 y +350 =78750,

Simplify constants:

-78400 +350 = -78050,

Thus,

15680x -280√14 y -78050 =78750,

Bring constants to the right:

15680x -280√14 y =78750 +78050 = 156800,

Divide both sides by 280:

15680x /280 = 56x,

-280√14 y /280 = -√14 y,

156800 /280 = 560.

Thus, simplified equation:

56x - √14 y = 560.

So that's the equation of the tangent at C: 56x - √14 y = 560.

Similarly, let me try to simplify the tangent at B equation.

Equation of tangent at B:

(-14/5)(x -5) + (311√14 /280)(y -5√14/56 ) = 78750/3136.

First, expand the terms:

-14/5 x +14/5 *5 + (311√14 /280)y - (311√14 /280)(5√14 /56 ) = 78750/3136.

Simplify term by term:

-14/5 x +14,

+ (311√14 /280)y,

- (311√14 *5√14 )/(280*56 )

Compute the last term:

Numerator: 311*5*(√14)^2 = 311*5*14 = 311*70 = 21770,

Denominator: 280*56 = 15680,

Thus, the term is -21770 /15680.

Therefore, the equation becomes:

-14/5 x +14 + (311√14 /280)y -21770/15680 = 78750/3136.

Multiply all terms by 15680 to eliminate denominators:

-14/5 x *15680 +14*15680 + (311√14 /280)y *15680 -21770 =78750/3136 *15680.

Compute each term:

First term: -14/5 x *15680 = -14*(15680/5)x = -14*3136x = -43904x,

Second term:14*15680 = 219520,

Third term: (311√14 /280)*15680 y = 311√14*(15680/280)y = 311√14*56 y = 311*56√14 y,

Fourth term: -21770,

Fifth term: 78750/3136 *15680 =78750*(15680/3136)=78750*5=393750.

Putting all together:

-43904x +219520 +311*56√14 y -21770 =393750,

Simplify constants:

219520 -21770 = 197750,

Thus,

-43904x +311*56√14 y +197750 =393750,

Bring constants to the right:

-43904x +311*56√14 y =393750 -197750 =196000,

Hmm, this is getting too complicated. Maybe there's a mistake here. Alternatively, perhaps I need to find another approach.

Alternatively, since both tangent equations are too complicated, maybe use parametric coordinates for point D?

Alternatively, since D is the exsymmedian point or something else?

Alternatively, use the concept that AP * AD = power of A with respect to some circle? Wait, but A is on ω, so power is zero. Hmm.

Wait, but power of D with respect to ω is DB² = DP * DA. So if I can compute DB and DA, then I can find AP.

Wait, AP = AD - DP, but DP = DB² / DA. Therefore, AP = AD - (DB²)/DA. Therefore, AP = (AD² - DB²)/AD. Hmm, so if I can compute AD and DB, then I can compute AP.

Therefore, if I can compute AD and DB, then AP can be found via that formula.

Therefore, perhaps compute DB and AD.

To compute DB and AD, perhaps use coordinates.

First, let me compute coordinates of D by solving the two tangent equations.

We have the tangent at C simplified to 56x - √14 y = 560.

Let me see if I can get the tangent at B into a similar simplified form.

Wait, perhaps I made an error in expanding the tangent at B equation. Let me try again.

Alternatively, let me see if I can find D as the intersection point of the two tangents. Since the tangent at C is 56x - √14 y = 560, and the tangent at B is another line. Let me see if I can find another equation for the tangent at B.

Alternatively, perhaps use parametric coordinates for D.

Alternatively, since D lies on both tangents, so it must satisfy both equations. Let me suppose that D has coordinates (d_x, d_y). Then, it must satisfy both tangent equations.

Alternatively, since the tangent at B and tangent at C meet at D, and DB = DC (since they are both tangent lengths from D to ω). Wait, yes! Since D is the external point from which two tangents are drawn to the circle, so DB = DC. Therefore, D lies on the perpendicular bisector of BC.

Given that BC is from point B (11/5, 6√14/5 ) to point C (10,0). The perpendicular bisector of BC would give the locus of points equidistant from B and C, so D lies on that.

Therefore, perhaps compute the perpendicular bisector of BC and find its intersection with the tangent at C (or B), which might be easier.

Let me compute the midpoint of BC first.

Coordinates of B: (11/5, 6√14/5 ), coordinates of C: (10,0).

Midpoint M of BC is:

x-coordinate: (11/5 +10)/2 = (11/5 +50/5)/2 = (61/5)/2 =61/10 =6.1,

y-coordinate: (6√14/5 +0)/2 = 3√14/5.

The slope of BC is (0 -6√14/5 )/(10 -11/5 ) = (-6√14/5 )/( (50/5 -11/5 )) = (-6√14/5 )/(39/5 ) = (-6√14)/39 = (-2√14)/13.

Therefore, the perpendicular bisector of BC has slope reciprocal and opposite: 13/(2√14 ).

Therefore, the equation of the perpendicular bisector is:

y - 3√14/5 = (13)/(2√14)(x -61/10 )

Since D lies on this perpendicular bisector and also on the tangent at C (56x -√14 y =560 ), we can solve these two equations to find D's coordinates.

Let me write the perpendicular bisector equation as:

y = (13)/(2√14)(x -61/10 ) + 3√14/5.

Let me substitute this into the tangent at C equation: 56x -√14 y =560.

Substitute y:

56x -√14 [ (13)/(2√14)(x -61/10 ) + 3√14/5 ] =560.

Simplify term by term:

First term: 56x,

Second term: -√14 * [13/(2√14)(x -61/10 ) ] = - [13/2 (x -61/10 ) ],

Third term: -√14*(3√14/5 ) = - (3*14)/5 = -42/5.

Therefore, the equation becomes:

56x - (13/2)(x -61/10 ) -42/5 =560.

Multiply all terms by 10 to eliminate denominators:

56x*10 - (13/2)(x -61/10 )*10 -42/5 *10 =560*10,

Compute each term:

560x -13*5(x -61/10 ) -84 =5600,

Simplify:

560x -65(x -6.1) -84 =5600,

Wait, 61/10 is 6.1,

So,

560x -65x +65*6.1 -84 =5600,

Compute 65*6.1:

65*6 =390, 65*0.1=6.5, total 396.5,

Thus,

560x -65x +396.5 -84 =5600,

Combine like terms:

(560 -65)x + (396.5 -84) =5600,

495x +312.5 =5600,

495x =5600 -312.5 =5287.5,

x =5287.5 /495,

Convert to fractions:

5287.5 =52875/10 =10575/2,

495 = 495/1,

Thus,

x = (10575/2)/495 =10575/(2*495)=10575/990.

Simplify numerator and denominator by dividing numerator and denominator by 15:

10575 ÷15 =705,

990 ÷15=66,

So 705/66. Again divide by 3:

705 ÷3=235,

66 ÷3=22,

Thus x=235/22 ≈10.6818.

Hmm, okay. Now compute y-coordinate using the perpendicular bisector equation:

y = (13)/(2√14)(x -61/10 ) + 3√14/5.

Plugging x =235/22,

First compute x -61/10:

235/22 -61/10 = (235*10 -61*22)/220 = (2350 -1342)/220 =1008/220 =504/110 =252/55.

Thus,

First term: (13)/(2√14)*(252/55 ) = (13*252)/(2√14 *55 ) = (3276)/(110√14 )

Second term: 3√14/5.

Therefore,

y = 3276/(110√14 ) + 3√14/5.

Let me rationalize the first term:

3276/(110√14 ) = (3276√14 )/(110*14 ) = (3276√14 )/1540.

Simplify numerator and denominator:

Divide numerator and denominator by 4:

3276/4 =819, 1540/4=385,

So 819√14 /385.

Second term: 3√14/5 = (3√14 *77)/385 = 231√14 /385.

Therefore,

y = (819√14 +231√14)/385 = (1050√14)/385.

Simplify:

1050 ÷35 =30, 385 ÷35=11,

Thus, y = (30√14)/11.

Therefore, coordinates of D are (235/22, 30√14/11 ).

Now that we have coordinates for A (0,0) and D (235/22, 30√14/11 ), we can find the equation of line AD and then find its other intersection point P with the circumcircle ω. Then compute the distance AP.

First, let me find parametric equations for line AD.

Parametrize AD from A(0,0) to D(235/22, 30√14/11 ). Let parameter t go from 0 to1 for segment AD, but since P is beyond A? Wait, no, P is on the circumcircle, so since A is on the circle, and AD goes through D, then P is the other intersection point, so actually, when extending AD beyond D, but wait, no. Wait, AD starts at A, goes through D, but since D is outside the circle, the line AD passes through A (on the circle) and then exits at P? Wait, no, actually, when you go from A towards D, since D is outside, the line AD starts at A, goes out to D, so the other intersection point P must be between A and D? Wait, but A is already on the circle, so the line AD starts at A, goes through the circle again at P? Wait, no, A is on the circle, so the line AD starts at A, which is on the circle, and then goes outward to D. Therefore, the other intersection point P must be the same as A? No, that can't be. Wait, perhaps I made a mistake here.

Wait, the problem states: "Let AD intersect ω at P". Since A is already on ω, then P is the other intersection point. Therefore, the line AD starts at A, goes through D, and intersects the circle again at P. Wait, but if D is outside the circle, then the line AD would pass through A (on the circle), go towards D (outside), so the other intersection point would be on the other side of the circle from D? Wait, no, actually, the line AD starts at A, goes through D, so the only other intersection with the circle would be on the opposite side of A from D. Wait, but that would require the line AD to pass through the circle again in the direction opposite to D, but since D is outside, the line from A through D would only intersect the circle at A and another point P between A and D? Wait, no, if D is outside, then the line AD would pass through A (on the circle) and then go to D outside, so the only other intersection is A. Therefore, perhaps the problem meant that AD is extended beyond A to intersect the circle again at P? That can't be, since A is already on the circle.

Wait, perhaps I made a mistake in the direction. Let me check the Asymptote code given:

The Asymptote code says:

draw(A..B..C..cycle, red+dotted);

O = circumcenter(A, B, C);

E = rotate(90,B) * (O); F = rotate(90,C) * (O);

D = IP(B..E + (B-E)*4, C..F + (C-F)*-3);

draw(B--D--C--D--A);

P = IP(D..A, A..B..C);

So, in the Asymptote code, P is the intersection point of line DA and the circumcircle, other than A. So, since D is outside the circle, the line DA passes through A (on the circle) and goes to D, so the other intersection point P must be the same as A? That can't be. Wait, perhaps the Asymptote code uses IP to find the other intersection, so maybe P is A? But the problem states "AD intersects ω at P", so probably P is the other intersection, so perhaps the line AD is extended beyond A to meet the circle again at P? But that would mean AP is a chord passing through A and P, with D lying on the extension beyond A. But in that case, the power of point D would be DP * DA = DB², where DP is the distance from D to P, and DA is from D to A. Wait, but if P is on the other side, then DA would be from D to A, and DP would be DA + AP. Hmm, perhaps.

Wait, let me clarify. Let me suppose that the line AD passes through A and goes to D, which is outside the circle. Then the two intersection points of line AD with the circle are A and another point P which is between A and D? Wait, but if D is outside, then the line from A to D would pass through A (on the circle) and then exit the circle at some point P between A and D? Wait, no, if D is outside, then the line AD would start at A (on the circle), go towards D (outside), so the only other intersection is A. Therefore, perhaps the problem has a typo, or I misunderstood the configuration.

Wait, looking back at the problem statement:

"Let the tangents to ω at B and C intersect at point D, and let AD intersect ω at P."

Since AD is a line from A to D, which is outside the circle, so the intersections with ω are A and another point P. Therefore, P must be the second intersection point, so the line AD must pass through A, then P, then D? Wait, but that would mean that P is between A and D. But then the power of point D with respect to ω is DP * DA = DB². Wait, yes, if P is between A and D, then DP is the distance from D to P, and DA is the distance from D to A, so DP * DA would be (DA - AP) * DA. Wait, but power of D is DB² = DP * DA, where DP is the entire length from D to P, and DA is from D to A? Wait, no, the power of a point D with respect to the circle is equal to the product of the distances from D to each intersection point along a line. So if the line AD passes through P and A, with P closer to D, then power of D is DP * DA, where DP is from D to P, and DA is from D to A. Wait, but A is on the circle, so DA is the distance from D to A, and DP is from D to P. Since P is between D and A, then DA = DP + PA. Therefore, power of D is DP * DA = DP*(DP + PA) = DB².

Therefore, if I can write equations for AP in terms of DP and DA, then perhaps solve for AP.

Alternatively, since we have coordinates for A and D, we can parametrize line AD and find its other intersection point P with the circumcircle.

Let me proceed with coordinates.

Coordinates of A: (0,0)

Coordinates of D: (235/22, 30√14/11 )

Parametrize line AD as follows:

Let parameter t be such that when t=0, we are at A(0,0), and t=1, we are at D(235/22, 30√14/11 ). The parametric equations are:

x = (235/22) * t,

y = (30√14/11 ) * t,

for t ∈ ℝ.

We need to find the other intersection point P of this line with the circumcircle ω. Since A is already on the circle (t=0), the other point P corresponds to some t ≠0.

The equation of the circumcircle ω is:

(x -5)^2 + (y -5√14/56 )^2 = (75√14 /56 )².

Plugging the parametric coordinates into the circle equation:

[ (235/22 t -5 ) ]² + [ (30√14/11 t -5√14/56 ) ]² = (75√14 /56 )².

This equation will have two solutions: t=0 (point A) and t = t_P (point P). Let me solve for t.

First, compute each term:

First term: (235/22 t -5 )²

Let me write 5 as 110/22 to have the same denominator:

235/22 t - 110/22 = (235t -110)/22,

Thus, first term squared: (235t -110)^2 / (22)^2.

Second term: (30√14/11 t -5√14/56 )².

Factor out √14:

√14*(30/11 t -5/56 ),

Thus squared: 14*(30/11 t -5/56 )².

Therefore, the second term squared is 14*( (30/11 t -5/56 ) )².

Therefore, the equation becomes:

[ (235t -110)^2 ] / (22² ) + 14*(30/11 t -5/56 )² = (75√14 /56 )².

This is quite complex, but let me compute each part step by step.

First, compute the right-hand side (RHS):

(75√14 /56 )² = (75² *14)/ (56² ) = (5625 *14)/3136 = 78750/3136.

Now, let me compute the first term on the left:

(235t -110)^2 / (22² ) = ( (235t -110)^2 ) / 484.

Second term:

14*( (30/11 t -5/56 )^2 )

Let me compute the expression inside the square:

30/11 t -5/56 = (30*56 t -5*11 )/(11*56 ) = (1680t -55)/616.

Therefore, squared term is (1680t -55)^2 / (616² ),

Thus, the second term becomes 14*( (1680t -55)^2 )/(616² ).

Therefore, the equation is:

(235t -110)^2 /484 + 14*(1680t -55)^2/(616² ) =78750/3136.

This is very algebra-heavy. Let me see if I can find a common denominator or simplify terms.

Alternatively, perhaps factor out terms:

Note that 616 = 22*28, and 484 =22²,

Let me see:

First term denominator: 22²,

Second term denominator: (616)^2 = (22*28)^2 =22² *28²,

Thus, the second term can be written as 14*(1680t -55)^2/(22² *28² )

Therefore, the entire equation can be multiplied by 22² *28² to eliminate denominators:

(235t -110)^2 *28² +14*(1680t -55)^2 = (78750/3136)*22² *28².

This is still very involved, but let me compute each part step by step.

First, compute (235t -110)^2 *28²:

Let me denote term1 = (235t -110)^2 *784 (since 28²=784),

term2 =14*(1680t -55)^2,

RHS: (78750/3136)*(484)*(784). Let me compute that:

First, note that 3136 = 56², 484=22², 784=28².

Compute 78750/3136 *484*784:

=78750 *484 *784 /3136,

Simplify denominators:

3136 =56² = (7*8)^2 =7²*8²,

784 =28² = (4*7)^2 =4²*7²,

484 =22²,

Thus,

78750 *22² * (4²*7²) / (7²*8²) )

Wait, let me compute step by step:

First, 784 /3136 = (28²)/(56²) = (28/56)^2 = (1/2)^2 =1/4,

Thus,

78750 *484 *784 /3136 =78750 *484*(784/3136)=78750 *484*(1/4)=78750 *121.

Wait, 784/3136 = 0.25, so 784/3136 =1/4,

Thus,

78750 *484 * (1/4) =78750 *121,

Because 484 divided by4 is 121.

Therefore, RHS =78750 *121.

Compute 78750 *121:

First compute 78750 *100 =7,875,000,

78750 *20 =1,575,000,

78750 *1 =78,750,

Total: 7,875,000 +1,575,000 =9,450,000 +78,750 =9,528,750.

Thus RHS =9,528,750.

Now, compute term1 + term2 =9,528,750.

This is going to be a quadratic equation in t. Let me compute term1 and term2.

First, compute term1 = (235t -110)^2 *784.

Let me expand (235t -110)^2:

= (235t)^2 - 2*235t*110 +110²

=55225t² -51700t +12100.

Multiply by 784:

term1 = (55225t² -51700t +12100)*784.

Similarly, term2 =14*(1680t -55)^2.

Expand (1680t -55)^2:

= (1680t)^2 -2*1680t*55 +55²

=2,822,400t² -184,800t +3,025.

Multiply by14:

term2 =14*(2,822,400t² -184,800t +3,025 )

=39,513,600t² -2,587,200t +42,350.

Therefore, term1 + term2:

First, compute term1:

55225t² *784 =55225*784 t²,

55225*784: Let me compute 55225*700 =38,657,500,

55225*84=55225*(80+4)=55225*80=4,418,000 +55225*4=220,900 → total 4,638,900,

Total term1 for t² term: 38,657,500 +4,638,900 =43,296,400 t²,

Similarly, -51700t *784 = -51700*784 t,

Compute 51700*700 =36,190,000,

51700*84=51700*(80+4)=41,360,000 +206,800 =41,566,800,

Total: 36,190,000 +41,566,800 =77,756,800,

Thus, the t term from term1 is -77,756,800 t,

Constant term from term1: 12100*784 = 12100*700 +12100*84 =8,470,000 +1,028,400 =9,498,400.

Adding term2:

term1 + term2:

t² terms:43,296,400 t² +39,513,600t² =82,810,000 t²,

t terms: -77,756,800 t -2,587,200 t =-80,344,000 t,

constants:9,498,400 +42,350 =9,540,750.

Thus, the equation is:

82,810,000 t² -80,344,000 t +9,540,750 =9,528,750,

Subtract RHS:

82,810,000 t² -80,344,000 t + (9,540,750 -9,528,750 )=0,

Which simplifies to:

82,810,000 t² -80,344,000 t +12,000 =0.

Divide all terms by 2000 to simplify:

41,405 t² -40,172 t +6 =0.

Wait, let me check:

82,810,000 /2000 =41,405,

80,344,000 /2000 =40,172,

12,000 /2000 =6.

Thus, quadratic equation:41,405 t² -40,172 t +6 =0.

Hmm, let me see if this can be simplified further. Let me check if coefficients have a common factor.

41,405 and 40,172: Let's see, 41,405 -40,172 =1,233.

Check GCD of 40,172 and 1,233.

40,172 ÷1,233 ≈32.6, 1,233*32=39,456, 40,172-39,456=716,

GCD(1,233,716):

1,233 ÷716 =1 with remainder 517,

716 ÷517=1 rem 199,

517 ÷199=2 rem 119,

199 ÷119=1 rem80,

119 ÷80=1 rem39,

80 ÷39=2 rem2,

39 ÷2=19 rem1,

2 ÷1=2 rem0. So GCD is1.

Thus, no common factors. So we need to solve 41,405 t² -40,172 t +6 =0.

Use quadratic formula:

t = [40,172 ±√(40,172² -4*41,405*6 )]/(2*41,405 )

Compute discriminant:

D =40,172² -4*41,405*6.

First compute 40,172²:

This is a big number. Let me see:

40,172² = (40,000 +172)^2 =40,000² +2*40,000*172 +172²

=1,600,000,000 +13,760,000 +29,584 =1,613,789,584.

Then compute 4*41,405*6 =4*6*41,405 =24*41,405 =993,720.

Thus, D =1,613,789,584 -993,720 =1,612,795,864.

Now compute sqrt(D). Let me see if this is a perfect square.

Let me see, 40,172² is 1,613,789,584, which is larger than D by 993,720. Hmm, so sqrt(D) is slightly less than 40,172.

Alternatively, perhaps factor D:

But this might be too time-consuming. Alternatively, proceed numerically.

Alternatively, perhaps there is a calculation mistake earlier. This is getting too cumbersome. Maybe there is a smarter approach.

Alternatively, since we have coordinates for A and D, compute vector AD, then find point P such that AP * AD = power of A, but A is on the circle so power is zero. Wait, no.

Wait, power of D is DB² = DP * DA.

We can compute DB and DA using coordinates, then solve for DP, then AP = AD - DP.

Let me try this approach.

First, compute DB:

Coordinates of D: (235/22, 30√14/11 )

Coordinates of B: (11/5, 6√14/5 )

Compute the distance DB:

Δx =235/22 -11/5 = (235*5 -11*22)/110 = (1175 -242)/110 =933/110,

Δy =30√14/11 -6√14/5 = (30√14 *5 -6√14 *11)/55 = (150√14 -66√14)/55 =84√14 /55,

Thus, DB² = (933/110)^2 + (84√14 /55 )^2.

Compute each term:

First term: (933)^2/(110)^2,

Second term: (84² *14)/(55² ).

Compute first term numerator: 933². Let me compute 933²:

900²=810,000,

2*900*33=59,400,

33²=1,089,

Total: 810,000 +59,400 +1,089 =870,489.

Thus first term:870,489 /12,100.

Second term numerator:84²=7056, so 7056*14=98,784,

Denominator:55²=3025,

Thus second term:98,784 /3025.

Therefore, DB² =870,489/12,100 +98,784/3025.

Convert to common denominator 12,100 (since 3025*4=12,100):

Second term:98,784/3025 = (98,784 *4)/12,100 =395,136/12,100,

Thus total DB² = (870,489 +395,136)/12,100 =1,265,625 /12,100.

Simplify:

1,265,625 ÷25 =50,625,

12,100 ÷25=484,

Thus DB² =50,625 /484.

Now compute DA:

Coordinates of D: (235/22, 30√14/11 ), A is (0,0).

DA² = (235/22)^2 + (30√14/11 )^2.

Compute:

First term: (235)^2/(22)^2 =55,225 /484,

Second term: (900*14)/121 =12,600 /121,

Thus DA² =55,225/484 +12,600/121.

Convert to common denominator 484:

12,600/121 = (12,600 *4)/484 =50,400/484,

Thus DA² = (55,225 +50,400)/484 =105,625 /484,

Therefore DA = sqrt(105,625 /484 ) = 325/22 (since 325²=105,625 and 22²=484).

Thus DA =325/22.

Now, using the power of point D: DB² = DP * DA,

We have DB² =50,625 /484,

DA =325/22,

Thus,

DP = DB² / DA = (50,625 /484 ) / (325/22 ) = (50,625 /484 ) * (22 /325 )

Simplify:

50,625 ÷325 = 50,625 ÷25 ÷13 =2,025 ÷13 ≈155.769, but let me compute exactly:

325*155 =325*(150+5)=48,750 +1,625=50,375,

50,625 -50,375=250,

Thus 50,625 =325*155 +250,

250=325*(250/325)=325*(10/13),

Thus 50,625 /325 =155 +10/13 =155.769...

But let me keep it as fractions:

50,625 /325 = (50,625 ÷25)/(325 ÷25)=2,025/13,

Similarly, 22/484 =11/242,

Wait, let me re-express:

DP = (50,625 /484 ) * (22 /325 ) = (50,625 *22 )/(484 *325 )

Simplify numerator and denominator:

Note that 484 =22², so 484=22*22,

Thus,

= (50,625 *22 )/(22*22 *325 )

Cancel one 22:

=50,625/(22*325 )

Compute 50,625 ÷325 =50,625 ÷ (25*13)= (50,625 ÷25)/13=2,025/13,

Thus,

DP = (2,025/13)/22 =2,025/(13*22)=2,025/286.

Therefore, DP =2025/286.

Then, AP = AD - DP = (325/22 ) - (2025/286 ).

Convert to common denominator 286:

325/22 = (325*13)/286 =4,225/286,

Thus,

AP =4,225/286 -2,025/286 = (4,225 -2,025)/286 =2,200/286.

Simplify 2,200/286:

Divide numerator and denominator by 2:1,100/143.

Check if 1,100 and143 have common factors:

143=11*13,

1,100 ÷11=100,

Thus, 1,100=11*100,

143=11*13,

Thus, common factor 11:

1,100/143 = (11*100)/(11*13)=100/13.

Therefore, AP=100/13.

Thus, m=100, n=13, which are coprime. Therefore, m +n=113.

**Final Answer**
\boxed{113}
</think>

Given triangle \(ABC\) inscribed in circle \(\omega\) with \(AB = 5\), \(BC = 9\), and \(AC = 10\). The tangents to \(\omega\) at \(B\) and \(C\) intersect at point \(D\), and line \(AD\) intersects \(\omega\) again at \(P\). We need to find \(AP\) in the form \(\frac{m}{n}\) and compute \(m + n\).

1. **Coordinates of Points**:
   - Place \(A\) at \((0, 0)\), \(C\) at \((10, 0)\).
   - Coordinates of \(B\) are found using the distance formula:
     \[
     B\left(\frac{11}{5}, \frac{6\sqrt{14}}{5}\right)
     \]

2. **Circumcircle of \(ABC\)**:
   - Midpoint of \(AC\) is \((5, 0)\), perpendicular bisector is \(x = 5\).
   - Center \(O\) of the circumcircle is found to be \(\left(5, \frac{5\sqrt{14}}{56}\right)\).

3. **Equation of Tangents**:
   - Tangent at \(B\) and \(C\) are derived using the point-form of the tangent equation.
   - Solving the equations of tangents at \(B\) and \(C\) gives coordinates of \(D\):
     \[
     D\left(\frac{235}{22}, \frac{30\sqrt{14}}{11}\right)
     \]

4. **Power of Point \(D\)**:
   - Using the power of point \(D\) with respect to \(\omega\):
     \[
     DB^2 = DP \cdot DA
     \]
   - Calculate \(DB\) and \(DA\):
     \[
     DB^2 = \frac{50625}{484}, \quad DA = \frac{325}{22}
     \]
   - Solve for \(DP\):
     \[
     DP = \frac{DB^2}{DA} = \frac{2025}{286}
     \]
   - Compute \(AP\):
     \[
     AP = DA - DP = \frac{325}{22} - \frac{2025}{286} = \frac{100}{13}
     \]

5. **Result**:
   - \(AP = \frac{100}{13}\), so \(m = 100\) and \(n = 13\). Thus, \(m + n = 113\).

\[
\boxed{113}
\]