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 $m + n$.<｜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, and then the line AD intersects the circle ω again at point P. We are given the sides AB = 5, BC = 9, and AC = 10. We need to find the length AP expressed as a fraction m/n where m and n are coprime, and then compute m + n.

Hmm, first, let me recall some concepts. Since D is the intersection of the tangents at B and C to the circumcircle ω of triangle ABC, D must be the exsimilicenter or the insimilicenter of the circle? Wait, actually, the intersection of two tangents to a circle from a common external point is that point. So, D is the external point from which two tangents are drawn to the circle ω, touching it at B and C. Therefore, D is the ex-pole or something? Maybe I need to recall that the point where two tangents meet is called the exsimilicenter if they are external tangents. But perhaps more importantly, properties related to pole and polar.

Alternatively, maybe using power of a point. The power of point D with respect to ω should be equal to DB² = DC², since both are tangent lengths from D to ω. So, DB = DC, which makes D lie on the perpendicular bisector of BC? Wait, but BC is a chord of the circle, so the perpendicular bisector of BC passes through the center of the circle. But D is outside the circle, so maybe not exactly on that bisector, but since DB and DC are equal, D must lie on the perpendicular bisector of BC. Hmm, that might be useful.

Alternatively, maybe using harmonic division, projective geometry, or inversion. But perhaps starting with coordinate geometry might be more straightforward for me. Let me see.

Alternatively, using triangle properties. Since we have the sides given, maybe we can compute some angles or use the Law of Cosines first?

Wait, let me note down the given sides: AB = 5, BC = 9, AC = 10. So triangle ABC has sides 5, 9, 10. Let me check if it's a valid triangle. 5 + 9 > 10? 14 > 10, yes. 5 + 10 > 9? 15 > 9. 9 + 10 > 5? 19 >5. So yes, valid.

Perhaps first compute some key elements like the circumradius? Since the triangle is inscribed in circle ω, so the circumradius R can be computed using the formula:

R = (a*b*c)/(4*A), where a, b, c are the sides, and A is the area.

First, let me compute the area using Heron's formula. The semi-perimeter s = (5 + 9 + 10)/2 = 12. Then area A = sqrt[s(s - a)(s - b)(s - c)] = sqrt[12*(12 -5)(12 -9)(12 -10)] = sqrt[12*7*3*2] = sqrt[504] = sqrt[16*31.5] wait, 504 factors: 16*31.5? Wait 16*31.5 is 504, but 31.5 is not integer. Let me factor 504 properly. 504 divided by 16 is 31.5, but perhaps better to factor into primes: 504 = 8*63 = 8*7*9 = 8*7*9 = 2^3 * 3^2 *7. So sqrt(2^3 *3^2 *7) = 2^(1.5)*3*sqrt(7) = 2*sqrt(2)*3*sqrt(7) = 6*sqrt(14). Wait, let me check again:

Wait Heron's formula: sqrt[12*7*3*2] = sqrt[(12)*(7)*(3)*(2)] = sqrt[(12*3)*(7*2)] = sqrt[36 *14] = 6*sqrt(14). Ah, yes, that's better. So area A = 6√14.

Therefore, the circumradius R = (a*b*c)/(4*A) = (5*9*10)/(4*6√14). Let me compute numerator: 5*9=45, 45*10=450. Denominator: 4*6=24, so 24√14. Thus R = 450/(24√14) = simplify numerator and denominator by 6: 75/(4√14). Rationalizing the denominator: 75/(4√14) * √14/√14 = 75√14/(4*14) = 75√14 /56. Hmm, so R = (75√14)/56. Not sure if that's useful yet, but maybe.

Alternatively, perhaps using coordinates. Let me try coordinate geometry. Let me place triangle ABC in coordinate plane in a convenient way.

Let me set point B at (0,0), point C at (9,0), since BC =9, lying along the x-axis. Then point A is somewhere in the plane. Given that AB =5 and AC=10, so coordinates of A can be found by solving the equations.

Let me denote coordinates: Let’s let B = (0,0), C = (9,0). Let’s let A = (x,y). Then distance from A to B is 5: sqrt(x² + y²) =5, so x² + y² =25. Distance from A to C is 10: sqrt((x -9)^2 + y²)=10, so (x -9)^2 + y² =100.

Subtract the first equation from the second: (x -9)^2 + y² - (x² + y²) = 100 -25 => x² -18x +81 + y² -x² - y² =75 => -18x +81 =75 => -18x = -6 => x = 1/3.

So x = 1/3. Then from first equation, x² + y² =25 => (1/3)^2 + y² =25 => y² =25 -1/9 =224/9 => y= ±√(224)/3 = ±(√(16*14))/3 = ±4√14/3. Since triangles are typically considered in upper half-plane, let’s take y positive. So coordinates of A are (1/3, 4√14/3).

Alright, so coordinates:

A: (1/3, 4√14/3)

B: (0,0)

C: (9,0)

Now, need to find the circumcircle ω of triangle ABC. Since we have coordinates for A, B, C, perhaps we can compute the equation of the circle.

General equation of a circle: x² + y² + Dx + Ey + F =0. Plugging in points B(0,0): 0 +0 +0 +0 + F =0 => F=0.

So equation becomes x² + y² + Dx + Ey =0.

Plugging in point C(9,0): 81 +0 +9D +0 =0 => 9D = -81 => D = -9.

Plugging in point A(1/3, 4√14/3):

( (1/3)^2 + (4√14/3)^2 ) + D*(1/3) + E*(4√14/3) =0.

Compute each term:

(1/9 + (16*14)/9 ) + (-9)*(1/3) + E*(4√14/3) =0.

Compute first part: (1 + 224)/9 = 225/9 =25.

Then, -9*(1/3) = -3.

So 25 -3 + (4√14/3)*E =0 => 22 + (4√14/3)*E =0 => (4√14/3)*E = -22 => E = (-22)*(3)/(4√14) = (-66)/(4√14) = (-33)/(2√14). Rationalizing denominator: (-33√14)/(2*14) = (-33√14)/28.

Therefore, the equation of the circle is x² + y² -9x + Ey =0, with E = -33√14/(28). Wait, but let me confirm:

Wait, actually, in the equation above, when plugging in point A, we had:

25 (from the first terms) plus (-9)*(1/3) which is -3, plus E*(4√14/3) equals zero. So 25 -3 + (4√14/3)E =0, so 22 + (4√14/3)E =0, so solving for E gives E = (-22)*(3)/(4√14) = -66/(4√14) = -33/(2√14). So yes, that's correct. So E is -33/(2√14). So the equation is x² + y² -9x - (33/(2√14)) y =0.

Hmm, this seems a bit messy, but maybe manageable. Alternatively, perhaps compute the circumcircle's center and radius.

The general equation can be written as (x - h)^2 + (y - k)^2 = R^2. Since we have F=0, the center is at (-D/2, -E/2). From the equation x² + y² + Dx + Ey + F =0, so here D coefficient is -9, so h = -D/2 = 9/2. Similarly, k = -E/2. Since E is -33/(2√14), so -E/2 = (33/(2√14))/2 = 33/(4√14). Therefore, the center is at (9/2, 33/(4√14)). Hmm, complicated coordinates. Maybe better to work with coordinates of points and parametrize line AD?

Alternatively, perhaps find point D first. Since D is the intersection of the two tangents at B and C to the circumcircle. To find D, we can use the fact that 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 of the circle. Similarly, tangent at C is perpendicular to OC.

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

Given that, let me first compute the coordinates of the center O. From above, the center is (h, k) = (9/2, 33/(4√14)). Let me compute that numerically to see if that helps, but maybe symbolically is better.

Alternatively, perhaps there is a better way. Let me think.

Alternatively, since D is the ex-pole of BC or something? Wait, maybe using polars. The polar of D is the line BC, since D is the intersection of the tangents at B and C. So, the polar of D is BC. Therefore, by La Hire's theorem, if D lies on the polar of some point, then that point lies on the polar of D. Not sure if that helps here.

Alternatively, since we have coordinates for B and C, and the center O, perhaps compute equations of the tangents at B and C.

The tangent at point B(0,0): The radius OB is the vector from O to B, which is ( -9/2, -33/(4√14) ). The tangent line at B is perpendicular to OB. So, the slope of OB is (k - 0)/(h -0) = (33/(4√14)) / (9/2) = (33/(4√14)) * (2/9) = (33)/(18√14) = 11/(6√14). Therefore, the slope of the tangent at B is the negative reciprocal, which is -6√14 /11.

Therefore, the tangent at B has slope m1 = -6√14 /11, and passes through B(0,0), so its equation is y = (-6√14 /11)x.

Similarly, the tangent at point C(9,0). The radius OC is the vector from O to C: (9 - 9/2, 0 - 33/(4√14)) = (9/2, -33/(4√14)). The slope of OC is ( -33/(4√14) - 0 ) / (9/2 -9) wait, no, coordinates of C are (9,0), so the vector from O to C is (9 - 9/2, 0 - 33/(4√14)) = (9/2, -33/(4√14)). Therefore, the slope of OC is [ -33/(4√14) ] / (9/2) = (-33/(4√14)) * (2/9) = (-33)/(18√14) = -11/(6√14). Therefore, the tangent at C is perpendicular to OC, so its slope is the negative reciprocal of that, which is (6√14)/11.

Therefore, the tangent at C has slope m2 = 6√14 /11, and passes through point C(9,0). So its equation is y -0 = (6√14 /11)(x -9).

So now, we have equations of the two tangents:

Tangent at B: y = (-6√14 /11)x.

Tangent at C: y = (6√14 /11)(x -9).

We need to find their intersection point D. Let me set the two equations equal:

(-6√14 /11)x = (6√14 /11)(x -9).

Multiply both sides by 11/(6√14) to simplify:

- x = (x -9)

=> -x = x -9

=> -2x = -9

=> x = 9/2.

Then plug back x =9/2 into tangent at B's equation: y = (-6√14 /11)*(9/2) = (-54√14)/22 = (-27√14)/11.

Therefore, point D is at (9/2, -27√14/11).

Hmm, okay. So coordinates of D are (9/2, -27√14/11). Now, need to find the equation of line AD, which connects point A(1/3, 4√14/3) and D(9/2, -27√14/11). Then find where this line intersects the circle ω again at point P, and compute the distance from A to P.

First, let me find the parametric equation of line AD. Let me denote parameter t such that when t=0, we are at A, and t=1, we are at D. But maybe better to use a parameter that allows us to find the intersection point P beyond D? Wait, actually, since AD is extended to meet the circle again at P, so P is on the line AD beyond D from A? Wait, actually, since A is already on the circle, so the line AD starts at A, goes through D, and then intersects the circle again at P. Wait, but D is outside the circle, so the line AD starts at A (on the circle), goes through D (outside), and then exits the circle again at P? Wait, but actually, since D is outside the circle, the line AD would intersect the circle again at another point P. Wait, but A is already on the circle, so AP is the segment from A to P, passing through D? Wait, no, actually, AD is from A to D, but since D is outside, the line AD would pass through A, go out to D, but the other intersection point with the circle would be P. Wait, but A is already on the circle, so the line AD intersects the circle at A and P, so P is the other intersection point. Therefore, AP is the length from A to P, which is the same as the length from A to D extended to P. Hmm, so perhaps using power of a point D with respect to ω?

Wait, power of point D with respect to ω is equal to DB² = DC² (since D is the external point from which tangents are drawn). Also, power of D is equal to DA * DP, since D lies on the secant line ADP. Therefore, DB² = DA * DP. Since we need AP, which is equal to AD + DP? Wait, but depending on the direction. Wait, actually, since P is on the other side of D from A, then AP = AD + DP? Wait, no, if A is on the circle, and D is outside, then the line AD passes through A, goes to D, and then P is on the other side of D? Wait, actually, no. Let me think again.

Wait, the line AD starts at A (on the circle), goes through D (outside the circle), and then the other intersection point P must be on the opposite side of D from A? Wait, but the line AD already starts at A, so the two intersection points with the circle are A and P. Since D is outside, then P must lie between A and D? Wait, no. Let me recall that the power of point D with respect to the circle is equal to DB² = DA * DP. Since D is outside, the power is equal to DA * DP, where DA is the length from D to A, and DP is from D to P. But since A is on the circle, then DA * DP = DB², so DP = DB² / DA. Therefore, AP would be AD + DP? Wait, no, actually, the points are arranged as A---D---P, so AP = AD + DP. Wait, but if P is on the same line beyond D, then yes. Wait, but let me confirm.

Wait, power of a point D: If a line through D intersects the circle at A and P, then DA * DP = power of D. But depending on the direction, the lengths are signed. But since we are dealing with magnitudes, it's |DA| * |DP| = power. But since D is outside, and A is closer to D than P, then actually, the power is equal to DA * DP, where DP is the entire length from D to P, and DA is from D to A. Wait, actually, the formula is: If from point D outside the circle, a secant DAP is drawn, intersecting the circle at A and P (with A closer to D), then DA * DP = power of D. Wait, actually, no, the correct formula is DA * DP = power, but DA is the distance from D to A, and DP is from D to P, so if A is closer, then DA * DP = (distance from D to A) * (distance from D to P). But since P is further away, then DP = DA + AP? Wait, maybe better to use directed lengths.

Alternatively, let me think in terms of coordinates. Let me compute the coordinates of point P by parametrizing line AD and finding its other intersection with the circle.

First, let me find parametric equations for line AD. Let me denote point A as (1/3, 4√14/3) and D as (9/2, -27√14/11). Let me compute the vector from A to D:

Δx = 9/2 - 1/3 = (27/6 - 2/6) = 25/6

Δy = (-27√14/11) - (4√14/3) = let's compute this:

First, common denominator 33:

= (-27√14/11)*(3/3) - (4√14/3)*(11/11)

= (-81√14/33) - (44√14/33)

= (-125√14)/33

Therefore, the direction vector from A to D is (25/6, -125√14/33). We can parametrize the line AD as:

x = 1/3 + (25/6)t

y = (4√14)/3 + (-125√14/33)t

Where t=0 corresponds to A, and t=1 corresponds to D. To find the other intersection point P, we need to find t such that (x,y) is on the circle ω, besides point A (t=0). So plug the parametric equations into the circle's equation.

The circle equation is x² + y² -9x - (33/(2√14))y =0. Let me substitute x and y:

First, compute x²:

[1/3 + (25/6)t]^2 = (1/3)^2 + 2*(1/3)*(25/6)t + (25/6 t)^2 = 1/9 + (50/18)t + (625/36)t²

Similarly, y²:

[ (4√14/3) + (-125√14/33)t ]^2

Let me factor out √14:

√14 [4/3 - (125/33)t ] , so squared:

14*(4/3 - (125/33)t)^2

Expand that:

14*[ (16/9) - 2*(4/3)*(125/33)t + (125/33)^2 t² ]

=14*(16/9 - (1000/99)t + (15625/1089)t² )

Then, the term -9x:

-9*[1/3 + (25/6)t] = -3 - (225/6)t = -3 - (75/2)t

The term - (33/(2√14))y:

- (33/(2√14)) * [4√14/3 + (-125√14/33)t ]

Simplify each part:

First term inside the brackets: 4√14/3, so multiplied by -33/(2√14):

- (33/(2√14))*(4√14/3) = - (33*4√14)/(2√14*3) )= - (132)/(6) = -22.

Second term: - (33/(2√14))*(-125√14/33)t = [ (33 *125√14 ) / (2√14 *33) ) ] t = (125/2) t.

Therefore, combining all terms:

The entire equation becomes:

[ x² + y² -9x - (33/(2√14))y ] =0

Breaking down each component:

x²: 1/9 + (50/18)t + (625/36)t²

y²: 14*(16/9 - (1000/99)t + (15625/1089)t² )

-9x: -3 - (75/2)t

- (33/(2√14))y: -22 + (125/2)t

Adding all these together equals zero.

This seems very algebra-heavy. Let me compute each part step by step.

First, compute x²:

1/9 + (50/18)t + (625/36)t²

Simplify fractions:

50/18 = 25/9,

625/36 remains.

So x² = 1/9 + (25/9)t + (625/36)t²

y²:

14*(16/9) = 224/9,

14*(-1000/99)t = -14000/99 t,

14*(15625/1089)t² = 218750/1089 t²

So y² = 224/9 - (14000/99)t + (218750/1089)t²

-9x term: -3 - (75/2)t

- (33/(2√14))y term: -22 + (125/2)t

Now, sum all terms:

Start with constants:

1/9 (from x²) + 224/9 (from y²) -3 (from -9x) -22 (from the last term)

Compute constants:

1/9 + 224/9 = 225/9 =25,

25 -3 =22,

22 -22 =0.

Next, terms with t:

From x²: (25/9)t,

From y²: -14000/99 t,

From -9x: -75/2 t,

From the last term: +125/2 t.

So total t terms:

(25/9 -14000/99 -75/2 +125/2 ) t

Let me compute each coefficient:

First, 25/9 -14000/99:

Convert to common denominator 99:

25/9 = 275/99,

So 275/99 -14000/99 = (275 -14000)/99 = -13725/99

Then, -75/2 +125/2 = (50)/2 =25.

So total t terms: (-13725/99 +25 ) t

Convert 25 to 2475/99:

25 = 2475/99,

Thus, -13725/99 +2475/99 = (-13725 +2475)/99 = (-11250)/99 = -11250/99 = simplify by dividing numerator and denominator by 9: -1250/11.

So the t terms contribute (-1250/11) t.

Now, terms with t²:

From x²: 625/36 t²,

From y²: 218750/1089 t²,

So total t² terms:

625/36 + 218750/1089.

Let me compute this:

Convert to common denominator. 36 and 1089. 1089 is 33²=1089, 36 is 6². LCM of 36 and 1089: 36 factors into 2²*3², 1089 is 3²*11². So LCM is 2²*3²*11² = 4*9*121 = 4*1089=4356.

Convert 625/36 to (625 * 121)/4356,

and 218750/1089 = (218750 *4)/4356.

Compute 625*121: 625*121. 625*100=62500, 625*20=12500, 625*1=625. So 62500+12500+625=75625.

Thus, 625/36 =75625/4356,

218750*4 = 875,000,

so 218750/1089 =875000/4356.

Therefore total t² terms: (75625 +875000)/4356 = 950625 /4356.

Simplify 950625 /4356:

Let me see if they have common factors. Let's divide numerator and denominator by 9:

950625 ÷9 = 105,625,

4356 ÷9 = 484.

So 105625 /484. Let me check if 105625 divides by 484. 484 is 22²=484. 105625 ÷25=4225, so 105625=25*4225=25*65². 484=4*121=4*11². Not sure if they have common factors. 105625 is 25*(65)^2, 484 is 4*(11)^2. So no common factors. So 105625/484 is the simplified fraction.

Therefore, the t² term is (105625/484) t².

Putting it all together, the equation becomes:

0*t² (from constants) + (-1250/11)t + (105625/484)t² =0.

Wait, actually, the entire expression equals zero, so:

(105625/484)t² - (1250/11)t =0.

Factor out t:

t*(105625/484 t - 1250/11 ) =0.

Therefore, solutions are t=0 (which corresponds to point A) and t = (1250/11)/(105625/484) ).

Compute t:

t = (1250/11) * (484/105625)

Simplify:

First, note that 484 = 4*121 =4*11²,

105625: Let's see, 105625 divided by 25 is 4225, which is 65². So 105625 =25*65² =25*(5*13)^2=25*25*13²=625*169. Hmm, but maybe factor numerator and denominator:

1250/11 * 484/105625

= (1250 *484)/(11 *105625)

Note that 484 /11 =44, so:

= (1250 *44)/105625

Simplify 1250/105625: 1250 divides by 25 is 50, 105625/25=4225.

So 50*44 /4225 = 2200 /4225.

Simplify 2200/4225: divide numerator and denominator by 25: 88/169.

Wait, let me check:

Wait 1250 *44 = 1250*40 +1250*4 =50,000 +5,000=55,000

Denominator: 11*105625 = 1,161,875? Wait, no, wait, actually, in the previous step, after substituting 484/11 =44, so denominator becomes 11*105625? Wait, no, original expression after substitution was (1250 *44)/ (105625). Because 484 divided by 11 is 44, so the denominator is 11*105625? Wait, no, original expression was (1250 *484)/(11 *105625). Since 484 divided by 11 is 44, so 484 =11*44, so substituting:

(1250 * 11 *44)/(11 *105625) = (1250 *44)/105625.

Yes, so 1250*44 = 55,000,

55,000 /105,625.

Divide numerator and denominator by 25: 55,000 /25 =2,200; 105,625/25=4,225.

2,200 /4,225. Divide numerator and denominator by 25 again: 2,200/25=88, 4,225/25=169.

So 88/169. Therefore, t=88/169.

Therefore, the parameter t for point P is t=88/169. Wait, but let me confirm:

Wait, 1250*44 = 1250*(40 +4)= 1250*40=50,000 +1250*4=5,000 → total 55,000.

Denominator 11*105,625: Wait, no, the denominator after substitution was 105,625, so 55,000 /105,625 = (55,000 ÷ 25)/(105,625 ÷25)= 2,200 /4,225, then ÷25 again gives 88/169. So yes, t=88/169.

Therefore, the coordinates of point P are given by plugging t=88/169 into the parametric equations:

x =1/3 + (25/6)*(88/169)

y = (4√14)/3 + (-125√14/33)*(88/169)

Let me compute x first:

Compute (25/6)*(88/169):

25*88 =2200,

6*169=1014,

So 2200/1014. Simplify: divide numerator and denominator by 2: 1100/507.

Thus x =1/3 + 1100/507.

Convert 1/3 to 169/507 (since 507=3*13*13, so 1/3 = 169/507? Wait 507 divided by 3 is 169. Yes, so 1/3 = 169/507.

Therefore, x =169/507 +1100/507 = (169 +1100)/507 =1269/507.

Simplify 1269/507: divide numerator and denominator by 3: 423/169. 423 divided by 13? 13*32=416, so 423-416=7, so no. So 423/169 is simplest.

Now compute y-coordinate:

First term: (4√14)/3,

Second term: (-125√14/33)*(88/169)

Compute the coefficient:

-125/33 *88/169 = -125*(88)/(33*169)

Note that 88/33 = 8/3, so:

-125*(8/3)/169 = -1000/(3*169) = -1000/507.

Thus the second term is (-1000/507)√14.

Therefore, y = (4√14)/3 - (1000√14)/507.

Convert 4/3 to 676/507 (since 4/3 = (4*169)/(3*169)=676/507),

Thus y = (676√14/507 -1000√14/507) = (-324√14)/507.

Simplify numerator and denominator:

-324 and 507: GCD(324,507). 507 ÷324 =1 with remainder 183,

324 ÷183=1 r.141,

183 ÷141=1 r.42,

141 ÷42=3 r.15,

42 ÷15=2 r.12,

15 ÷12=1 r.3,

12 ÷3=4 r.0. So GCD is 3.

Thus, -324/507 = -108/169.

Therefore, y = (-108√14)/169.

Therefore, coordinates of P are (423/169, -108√14/169).

Now, we need to compute the distance AP between point A (1/3,4√14/3) and P (423/169, -108√14/169).

Let me compute the differences in coordinates:

Δx = 423/169 -1/3 = (423*3 -169)/ (169*3) = (1269 -169)/507 = 1100/507.

Δy = (-108√14/169) - (4√14/3) = let's compute:

Common denominator 507 (since 169*3=507):

= (-108√14 *3)/507 - (4√14 *169)/507

= (-324√14 - 676√14)/507

= (-1000√14)/507.

Therefore, the distance AP is sqrt[(Δx)^2 + (Δy)^2]

Compute (Δx)^2 = (1100/507)^2,

(Δy)^2 = ( -1000√14 /507 )^2 = (1000^2 *14)/507^2.

Thus,

AP^2 = [1100² + 1000² *14 ] / (507²)

Let me compute numerator:

First compute 1100²: 1,210,000,

1000² *14 = 1,000,000 *14 =14,000,000,

Total numerator: 1,210,000 +14,000,000 =15,210,000.

Therefore, AP^2 =15,210,000 / (507²).

Take square root: AP = sqrt(15,210,000)/507.

sqrt(15,210,000) = sqrt(15,210,000). Let me see:

15,210,000 =15,210 *1000 = (1521 *10) * (1000) =1521 *10,000. sqrt(1521)=39, sqrt(10,000)=100. So sqrt(1521*10,000)=39*100=3900.

Therefore, sqrt(15,210,000)=3900.

Thus AP =3900 /507.

Simplify 3900/507:

Divide numerator and denominator by 3: 1300/169.

Check if 1300 and 169 have common factors. 169 is 13². 1300 ÷13=100. So 1300=13*100, 169=13². Thus,

1300/169 = (13*100)/(13*13) )=100/13.

Therefore, AP=100/13.

Wait, that's nice! So AP is 100/13, so m=100, n=13, which are coprime. Thus m +n=113.

Wait, let me confirm the calculation for AP^2:

Wait, Δx was 1100/507, so squared is (1100)^2/(507)^2,

Δy was (-1000√14)/507, so squared is (1000^2 *14)/(507)^2,

Thus total numerator: 1100² +1000²*14.

1100² = (11^2)*(100)^2 = 121*10,000=1,210,000,

1000²*14 =1,000,000*14=14,000,000,

Total numerator: 1,210,000 +14,000,000 =15,210,000,

sqrt(15,210,000)= sqrt(15,210,000). Let me confirm sqrt(15,210,000):

sqrt(15,210,000) = sqrt(15,210,000) = sqrt(15210000) = sqrt(1521 *10000) = sqrt(1521)*sqrt(10000)= 39*100=3900. Yes, correct.

Thus AP = 3900 /507. Then simplifying:

Divide numerator and denominator by 3: 3900/3=1300, 507/3=169.

1300/169. Since 169*7=1183, 169*7.692≈1300. Wait, but 169*7=1183, 1300-1183=117, which is 169*0.7, but actually, 169*100/13 =1300. Wait, 169 is 13², so 1300 divided by 169 is (1300/13)/13 =100/13. Yes, exactly. So 1300/169 =100/13. So AP=100/13. So the answer is 100 +13=113.

Wait, that seems straightforward in the end. Let me just check if there was a simpler way to get here, perhaps using power of a point.

Earlier, I thought of using power of point D: DB² = DA * DP.

If I can compute DB, DA, then can solve for DP, then AP = AD + DP? Wait, but need to be careful with signs.

First, compute DB. Since D is at (9/2, -27√14/11), and B is at (0,0). So distance DB is sqrt( (9/2)^2 + ( -27√14/11)^2 )

Compute (9/2)^2 =81/4,

(27√14 /11)^2 = (27² *14)/121 = (729*14)/121 = 10,206/121,

Thus DB² =81/4 +10,206/121.

To add these, common denominator 484 (4*121):

81/4 = (81*121)/484 = 9801/484,

10,206/121 = (10,206*4)/484 =40,824/484,

Thus DB² = (9801 +40,824)/484 =50,625/484.

So DB² =50,625/484.

Now compute DA: distance from D to A.

Coordinates of D: (9/2, -27√14/11),

Coordinates of A: (1/3,4√14/3).

Δx =9/2 -1/3 =25/6,

Δy= -27√14/11 -4√14/3 = same as before, which was -125√14/33.

Thus DA² = (25/6)^2 + (-125√14/33)^2.

Compute:

(25/6)^2 =625/36,

(-125√14/33)^2 = (125² *14)/33² = (15,625 *14)/1,089 =218,750/1,089.

Thus DA² =625/36 +218,750/1089.

Convert to common denominator 36*1089=39,204.

625/36 = (625*1089)/39,204,

218,750/1089 = (218,750*36)/39,204.

Compute numerator:

625*1089: 625*(1000 +89)=625,000 +55,625=680,625,

218,750*36: 218,750*30=6,562,500; 218,750*6=1,312,500 → total 7,875,000.

Total numerator:680,625 +7,875,000 =8,555,625.

Thus DA² =8,555,625 /39,204.

Simplify:

Divide numerator and denominator by 3:

8,555,625 ÷3=2,851,875,

39,204 ÷3=13,068.

Again divide by 3:

2,851,875 ÷3=950,625,

13,068 ÷3=4,356.

Again divide by 3:

950,625 ÷3=316,875,

4,356 ÷3=1,452.

Again divide by 3:

316,875 ÷3=105,625,

1,452 ÷3=484.

So now we have 105,625 /484. So DA² =105,625/484.

Therefore DA = sqrt(105,625/484)= sqrt(105,625)/sqrt(484)= 325/22. Wait sqrt(105,625): let me check 325²=105,625, yes. sqrt(484)=22. So DA=325/22.

Therefore, DA =325/22.

Power of point D: DB² = DA * DP,

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

Simplify:

50,625 ÷325: Let's see 325*150=48,750, 50,625-48,750=1,875,

325*5.769≈1,875. Wait, 325*5.769≈1,875. Alternatively, 50,625 ÷325:

Divide numerator and denominator by 25: 50,625/25=2,025; 325/25=13.

So 2,025/13 = 155.769. Wait, but 13*155=2015, 2025-2015=10, so 155 +10/13=155.769.

Alternatively, keep as fractions:

50,625/325 = (50,625 ÷25)/(325 ÷25)=2025/13.

Similarly, 22/484 =11/242= but wait, the expression is (50,625/484)*(22/325)= (50,625 *22)/(484 *325).

Note that 484=22², so 484=22*22,

Thus,

(50,625 *22)/(22² *325)=50,625/(22*325).

Compute denominator:22*325=7,150,

So 50,625 /7,150.

Simplify:

Divide numerator and denominator by 25: 50,625/25=2,025; 7,150/25=286.

2,025/286. Let me see if they can be divided by 13: 286 ÷13=22, 2025 ÷13≈155.7, not integer. 2025 divided by 11? 2025/11≈184.09. Not integer. So perhaps it's 2025/286.

Wait, but 2025=45², 286=2×11×13. Not sure. Anyway, DP=2025/286.

Wait, but let me check:

Wait, 50,625 divided by 325 is 50,625 /325 = 155.769...

Wait, but let me see:

Wait, 50,625 divided by 325:

325 × 150 = 48,750,

50,625 -48,750 =1,875,

325 ×5.769=1,875,

So total 155.769, which is 155 + 1875/325 =155 + 75/13=155 +5 +10/13=160 +10/13=160.769...

Wait, but perhaps better to keep as fractions.

Wait, but let me see:

Wait, DP = (DB²)/DA = (50625/484)/(325/22) = (50625/484)*(22/325) = (50625 *22)/(484*325).

Note that 484 is 22², so 484=22*22,

Thus, 22/(22*22)=1/22,

Thus, the expression becomes (50625)/(22*325).

Compute 50625 divided by 325:

50625 ÷325: 325×155= 325×100 +325×50 +325×5=32,500+16,250+1,625=50,375,

50,625-50,375=250,

So 155 +250/325=155 +10/13,

Thus, 50625/325=155 +10/13= (155×13 +10)/13=(2015 +10)/13=2025/13,

Therefore, DP= (2025/13)/22=2025/(13×22)=2025/286.

Thus DP=2025/286.

Therefore, AP = AD + DP? Wait, no. Wait, in power of a point, the formula is DB² = DA * DP, where DA is the distance from D to A, and DP is from D to P. Since A is between D and P, then AP = AD + DP? Wait, actually, if D is outside, and the secant is D-A-P, then DA is from D to A, and DP is from D to P, so AP = DP - DA.

Wait, let me think carefully.

If the points are arranged as D---A---P, then DA is the distance from D to A, DP is from D to P, so AP = DP - DA.

But according to the power of a point formula: DA * DP = DB²,

Wait, but actually, the power of point D is equal to DA * DP, where DA and DP are signed lengths if considering direction, but as magnitudes, it's DA * DP = DB².

But in terms of distances:

If A is between D and P, then DP = DA + AP,

Thus, DA*(DA + AP) = DB²,

But we need to find AP. Let me denote AP as x,

Then DA*(DA +x)=DB²,

Thus x = (DB² / DA ) - DA,

So AP = (DB² / DA ) - DA = (DP) - DA,

Which is consistent.

Therefore, AP = DP - DA,

But DP = DB² / DA,

Thus AP = (DB² / DA ) - DA = (DB² - DA²)/DA.

Alternatively, compute AP = DP - DA,

We have DP =2025/286,

DA =325/22,

Convert DA to denominator 286: 325/22 = (325*13)/(22*13)=4225/286,

Thus DP - DA =2025/286 -4225/286 = (2025 -4225)/286 = (-2200)/286,

Wait, that gives a negative value, which can't be. Hmm, so perhaps my arrangement is wrong.

Wait, perhaps the points are arranged as A---D---P, so that AP = AD + DP,

But in that case, the power of point D would be DA * DP = DB²,

Wait, but DA is the distance from D to A, which would be DA = distance from D to A,

and DP is the distance from D to P, so if P is further away, then DP = DA + AP,

Wait, so AP = DP - DA,

But then DA * DP = DA*(DA + AP) = DB²,

So AP = (DB² / DA ) - DA,

Which would be positive if DB² / DA > DA,

Which is true since DB² / DA = DP,

and DP is longer than DA,

Wait, but in our calculation,

DB² =50625/484 ≈104.6,

DA =325/22≈14.77,

Thus DB² / DA ≈104.6 /14.77≈7.08,

But DA is ≈14.77,

So 7.08 -14.77 would be negative, which is impossible.

Hmm, so there must be an error in the direction.

Wait, perhaps I have the direction reversed. Maybe P is between A and D?

Wait, but A is on the circle, and D is outside, so the line AD starts at A (on the circle), goes through D (outside), and then P is the other intersection point, which would be on the extension beyond D? Wait, no, the line AD starts at A, goes through D, and since D is outside, the other intersection point P must be on the other side of A? Wait, no, the line AD passes through A and D, with A on the circle and D outside. The other intersection point P must be on the opposite side of D from A? Wait, no, actually, when you have a point D outside the circle, and you draw a line through D intersecting the circle at A and P, then A and P are on opposite sides of D? Wait, no, actually, both points A and P are on the same line, with D outside, so one is between D and the circle, and the other is on the extension. Wait, actually, if the line passes through D and intersects the circle at two points, then one is closer to D (A) and the other is further away (P). Wait, but in our case, A is already on the circle, so when we draw line AD, which starts at A and goes to D, then the other intersection point P must be on the other side of A from D? Wait, that can't be, because then D is outside, and the line AD would only intersect the circle at A, unless extended beyond A, but then P would be on the opposite side of A from D.

Wait, perhaps my parametrization was incorrect. Let me think again.

In the parametrization earlier, when I set t=0 at A and t=1 at D, then the other intersection point P corresponds to a negative t value?

Wait, in the parametrization earlier, when solving for t, we had t=0 (point A) and t=88/169≈0.52, which is between 0 and1. Wait, that would mean that P is between A and D? But that contradicts the power of a point.

Wait, but according to the parametrization, when t=88/169≈0.52, which is between 0 and1, so P is between A and D. But then the power of point D would be DA * DP, where DA is the distance from D to A (which is DA=325/22≈14.77), and DP is the distance from D to P, which would be less than DA, so DA * DP would be less than DA², but DB² is larger than DA², so that can't be.

Hmm, so there must be a mistake here. Let me check the parametrization again.

Wait, in the parametrization, I set:

x =1/3 + (25/6)t,

y =4√14/3 + (-125√14/33)t,

where t=0 is A, t=1 is D.

When I solved for t, I got t=88/169≈0.52, which is between 0 and1, so that would place P between A and D. But according to power of a point, that would give DA * DP = (distance from D to A) * (distance from D to P). If P is between A and D, then DP = DA - AP,

Wait, let me denote AP as the distance from A to P,

then DP = DA - AP,

Thus, DA*(DA - AP) = DB²,

So DA² - DA*AP = DB²,

Thus, -DA*AP = DB² - DA²,

AP = (DA² - DB²)/DA,

But in our case, DA≈14.77, DB≈sqrt(50625/484)=sqrt(104.6)=≈10.23,

Wait, wait, DB squared is 50625/484≈104.6, so DB≈10.23,

Wait, but DA is 325/22≈14.77,

So DA is longer than DB,

Thus DA² - DB² is positive,

So AP=(DA² - DB²)/DA,

Compute DA²: (325/22)^2 =105,625/484,

DB²=50,625/484,

Thus DA² - DB² = (105,625 -50,625)/484 =55,000/484,

Thus AP= (55,000/484)/(325/22)= (55,000/484)*(22/325)= (55,000*22)/(484*325).

Simplify:

55,000 ÷325 = 55,000 ÷ 325 = 55,000 ÷ (25×13)= 2,200 ÷13≈169.23,

But let me compute:

55,000*22 =1,210,000,

484*325=157,300,

Thus AP=1,210,000 /157,300 =1,210,000 ÷157,300,

Divide numerator and denominator by 100:12,100 /1,573,

Divide numerator and denominator by 13: 12,100 ÷13≈930.77, 1,573 ÷13=121,

Wait 13×121=1,573,

12,100 ÷13= 930.769,

Wait, but 12,100 divided by13:

13×930=12,090,

12,100-12,090=10,

So 930 +10/13,

Thus AP= (930 +10/13)/121 = (12,100/13)/121 =12,100/(13×121)=12,100/(1,573).

Wait, but 12,100 divided by 121 is 100,

Thus 12,100/(13×121)=100/13,

Ah! So AP=100/13. Which matches the previous result.

Therefore, even though the parametrization gave t=88/169 between 0 and1, which would suggest P is between A and D, but according to the power of a point, AP is 100/13≈7.69, while DA is≈14.77,

So AP is shorter than DA, so P is between A and D. But according to the problem statement, AP is the length from A to P where P is the other intersection point. So even if P is between A and D, that's okay. The problem didn't specify that P is beyond D.

Wait, but in the problem statement: "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 through D, but since D is outside the circle, the line AD starting at A goes through D and then exits the circle again at P? Wait, no, if D is outside, then the line AD starting at A (on the circle) goes towards D (outside), so the other intersection point would be on the other side of A? Wait, no, the line AD passes through A and D. Since A is on the circle and D is outside, the line AD will intersect the circle only at A, unless extended beyond A. Wait, but the problem says "AD intersects ω at P". Since A is already on ω, then P must be the other intersection point, which would require extending AD beyond D? Wait, no, if you go from A through D, you leave the circle at A, so to intersect again, you have to go the other way, opposite to D.

Wait, perhaps there was a mistake in the parametrization direction.

Wait, in my parametrization, I set t=0 at A and t=1 at D, so the line beyond D would be t>1, but the other intersection point P is actually in the opposite direction, t negative?

Wait, but when solving the quadratic equation, we got t=0 and t=88/169. So only two intersection points are A (t=0) and P at t=88/169. That suggests that the line AD only intersects the circle at A and P, with P between A and D. So the problem statement must mean that P is the other intersection point, which is between A and D. So AP is indeed 100/13, which is approximately 7.69, which is less than AD≈14.77, so that makes sense.

Therefore, the answer is 100/13, so m +n=113.

Therefore, after going through coordinate geometry and verifying with power of a point, I think the answer is 113.

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

Given triangle \(ABC\) with sides \(AB = 5\), \(BC = 9\), and \(AC = 10\), we need to find the length \(AP\) where \(P\) is the second intersection point of line \(AD\) with the circumcircle \(\omega\) of \(ABC\). Here, \(D\) is the intersection of the tangents to \(\omega\) at \(B\) and \(C\).

1. **Coordinates of Points**:
   - Place \(B\) at \((0, 0)\) and \(C\) at \((9, 0)\).
   - Coordinates of \(A\) are found using the distance formula:
     \[
     A = \left(\frac{1}{3}, \frac{4\sqrt{14}}{3}\right)
     \]

2. **Equation of the Circumcircle**:
   - The circumcircle \(\omega\) has center \(\left(\frac{9}{2}, \frac{33}{4\sqrt{14}}\right)\) and radius \(R = \frac{75\sqrt{14}}{56}\).

3. **Finding Point \(D\)**:
   - Tangents at \(B\) and \(C\) intersect at \(D\). Using the slopes of the tangents and solving their equations, we find:
     \[
     D = \left(\frac{9}{2}, -\frac{27\sqrt{14}}{11}\right)
     \]

4. **Parametric Equation of Line \(AD\)**:
   - Parametrize line \(AD\) from \(A\) to \(D\):
     \[
     x = \frac{1}{3} + \frac{25}{6}t, \quad y = \frac{4\sqrt{14}}{3} - \frac{125\sqrt{14}}{33}t
     \]
   - Solving for intersection with the circumcircle, we find parameter \(t = \frac{88}{169}\).

5. **Coordinates of \(P\)**:
   - Substituting \(t = \frac{88}{169}\) into the parametric equations gives coordinates of \(P\):
     \[
     P = \left(\frac{423}{169}, -\frac{108\sqrt{14}}{169}\right)
     \]

6. **Distance \(AP\)**:
   - Calculate the distance between \(A\) and \(P\):
     \[
     AP = \sqrt{\left(\frac{1100}{507}\right)^2 + \left(\frac{-1000\sqrt{14}}{507}\right)^2} = \frac{100}{13}
     \]

7. **Final Result**:
   - The fraction \(\frac{100}{13}\) gives \(m = 100\) and \(n = 13\), so \(m + n = 113\).

\[
\boxed{113}
\]