Excircle Tangency Points

Geometry Level 5

Triangle A B C ABC has B C = 13 , C A = 14 , A B = 15. BC = 13, CA = 14, AB = 15. The C C -excircle of A B C \triangle ABC is tangent to the extensions of A C AC and B C BC at A A' and B , B', respectively, and is tangent to A B AB at D . D. Now, extend A D A'D to meet B C BC at X X and, similarly, extend B D B'D to meet A C AC at Y . Y. Letting P P be the point where A X AX and B Y BY intersect, the ratio C P P D \frac{CP}{PD} is equal to p q , \frac{p}{q}, where p p and q q are coprime positive integers.

Find p + q . p + q.


The answer is 53.

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2 solutions

Steven Yuan
Sep 1, 2017

(Geogebra couldn't let me input A A' and B , B', so I replaced them with A1 and B1 in the diagram.)

This is going to be a very long solution. If you have a shorter solution than this, then please post it!

For simplicity, let A = B A C , B = A B C . A = \angle BAC, B = \angle ABC. Also, we will let B C = a , C A = b , A B = c , BC = a, CA = b, AB = c, and s = a + b + c 2 . s = \frac{a + b + c}{2}.

First, we'll prove this lemma:

Lemma 1: A B A B = 2 cos A 2 cos B 2 . \dfrac{A'B'}{AB} = 2 \cos \dfrac{A}{2} \cos \dfrac{B}{2}.

Consider a general acute triangle A B C , ABC, and let the C C -excircle with center E E be tangent at A , B , D A', B', D as shown in the diagram.

Let I I be the incenter of A B C . \triangle ABC. Note that

B A D = 1 2 B E D = 1 2 ( 180 D B B ) = B 2 = I B A . \angle B'A'D = \dfrac{1}{2} \angle B'ED = \dfrac{1}{2} (180 - \angle DBB') = \dfrac{B}{2} = \angle IBA.

Similarly, A B D = A 2 = I A B . \angle A'BD = \dfrac{A}{2} = \angle IAB. Therefore, A B D B A I . \triangle A'B'D \sim \triangle BAI.

We can find that A D = 2 ( s a ) cos A 2 A'D = 2(s - a) \cos \frac{A}{2} and B I = s a cos B 2 , BI = \frac{s - a}{\cos \frac{B}{2}}, so

A B A B = A D B I = 2 ( s a ) cos A 2 s a cos B 2 = 2 cos A 2 cos B 2 . \dfrac{A'B'}{AB} = \dfrac{A'D}{BI} = \dfrac{2(s - a) \cos \frac{A}{2}}{\frac{s - a}{\cos \frac{B}{2}}} = 2 \cos \dfrac{A}{2} \cos \dfrac{B}{2}. \, \blacksquare

Next, we will find B X X C \frac{BX}{XC} and A Y Y C . \frac{AY}{YC}. To accomplish this, we will find the lengths of B X , X C , A Y , Y C B'X, XC, A'Y, YC via the Ratio Lemma, then use the facts that B X = B X B B BX = B'X - B'B and A Y = A Y A A . AY = A'Y - A'A.

Before moving on, we note that C A X = A 2 \angle CA'X = \frac{A}{2} and C B Y = B 2 . \angle CB'Y = \frac{B}{2}.

Lemma 2: B X X C = a + b + c a , A Y Y C = a b + c b \dfrac{B'X}{XC} = \dfrac{-a + b + c}{a}, \dfrac{A'Y}{YC} = \dfrac{a - b + c}{b}

By the results of Lemma 1, we have A B = 2 c cos A 2 cos B 2 . A'B' = 2c\cos \frac{A}{2} \cos \frac{B}{2}. Now, applying the Ratio Lemma for B X X C \frac{B'X}{XC} in A B C , \triangle A'B'C, we have

B X X C = A B A C sin B A X sin C A X = 2 c cos A 2 cos B 2 s sin B 2 sin A 2 = c cos A 2 sin B s sin A 2 . \begin{aligned} \dfrac{B'X}{XC} &= \dfrac{A'B'}{A'C} \cdot \dfrac{\sin \angle B'A'X}{\sin \angle CA'X} \\ &= \dfrac{2c\cos \frac{A}{2} \cos \frac{B}{2}}{s} \cdot \dfrac{\sin \frac{B}{2}}{\sin \frac{A}{2}} \\ &= \dfrac{c\cos \frac{A}{2} \sin B}{s\sin \frac{A}{2}}. \end{aligned}

We know that cos A 2 = s ( s a ) b c \cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}} and sin A 2 = ( s b ) ( s c ) b c \sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}} :

c cos A 2 sin B s sin A 2 = c sin B s s ( s a ) ( s b ) ( s c ) = c ( s a ) sin B s ( s a ) ( s b ) ( s c ) . \begin{aligned} \dfrac{c\cos \frac{A}{2} \sin B}{s\sin \frac{A}{2}} &= \dfrac{c \sin B}{s} \sqrt{\dfrac{s(s - a)}{(s - b)(s - c)}} \\ &= \dfrac{c(s - a) \sin B}{\sqrt{s(s - a)(s - b)(s - c)}}. \end{aligned}

We recognize the denominator as [ A B C ] , [ABC], by Heron's formula. By another area formula, we also have sin B = 2 [ A B C ] a c . \sin B = \frac{2 [ABC]}{ac}. Thus,

B X X C = c ( s a ) sin B s ( s a ) ( s b ) ( s c ) = c ( s a ) 2 [ A B C ] a c [ A B C ] = 2 ( s a ) a = a + b + c a . \begin{aligned} \dfrac{B'X}{XC} &= \dfrac{c(s - a) \sin B}{\sqrt{s(s - a)(s - b)(s - c)}} \\ &= \dfrac{c(s - a) \frac{2 [ABC]}{ac}}{[ABC]} \\ &= \dfrac{2(s - a)}{a} \\ &= \dfrac{-a + b + c}{a}. \end{aligned}

Using the same strategy, we can find A Y Y C = a b + c b . \frac{A'Y}{YC} = \frac{a - b + c}{b}. \, \blacksquare

Now, we can find B X X C \frac{BX}{XC} and A Y Y C \frac{AY}{YC} :

Lemma 3: B X X C = s a s , A Y Y C = s b s \dfrac{BX}{XC} = \dfrac{s-a}{s}, \dfrac{AY}{YC} = \dfrac{s-b}{s}

Again, we'll focus solely on the points and segments on ray C B . CB. By Lemma 2, B X = a + b + c a X C . B'X = \frac{-a + b + c}{a}XC. We also have B X + X C = s , B'X + XC = s, by the properties of the excircle. Thus,

B X + X C = s X C ( a + b + c a + 1 ) = s X C ( b + c a ) = s X C = s a b + c . B X = s s a b + c . \begin{aligned} B'X + XC &= s \\ XC \left ( \dfrac{-a + b + c}{a} + 1 \right ) &= s \\ XC \left ( \dfrac{b + c}{a} \right ) &= s \\ XC &= \dfrac{sa}{b + c}. \\ B'X &= s - \dfrac{sa}{b + c}. \end{aligned}

Next, we apply B X = B X B B BX = B'X - B'B :

B X = B X B B = s s a b + c ( s a ) = a s a b + c = a ( 1 a + b + c 2 ( b + c ) ) = a ( a + b + c ) 2 ( b + c ) = a ( s a ) b + c . \begin{aligned} BX &= B'X - B'B \\ &= s - \dfrac{sa}{b + c} - (s - a) \\ &= a - \dfrac{sa}{b + c} \\ &= a \left ( 1 - \dfrac{a + b + c}{2(b + c)} \right ) \\ &= \dfrac{a(-a + b + c)}{2(b + c)} \\ &= \dfrac{a(s - a)}{b + c}. \end{aligned}

Finally,

B X X C = a ( s a ) b + c s a b + c = s a s . \dfrac{BX}{XC} = \dfrac{\frac{a(s - a)}{b + c}}{\frac{sa}{b + c}} = \dfrac{s - a}{s}.

In a similar vein, we can find that A Y Y C = s b s . \frac{AY}{YC} = \frac{s-b}{s}. \, \blacksquare

Here's the last observation before we can calculate our final answer:

Lemma 4: C , P , D C, P, D are collinear.

Suppose C P CP intersects A B AB at point Z . Z. By Ceva's Theorem, Z B Z A X B X C Y C Y A = 1. \frac{ZB}{ZA} \cdot \frac{XB}{XC} \cdot \frac{YC}{YA} = 1. Applying the results of Lemma 3 yields

Z B Z A X B X C Y C Y A = 1 Z B Z A s a s s s b = 1 Z B Z A = s b s a . \begin{aligned} \frac{ZB}{ZA} \cdot \frac{XB}{XC} \cdot \frac{YC}{YA} &= 1 \\ \dfrac{ZB}{ZA} \cdot \frac{s - a}{s} \cdot \dfrac{s}{s - b} &= 1 \\ \dfrac{ZB}{ZA} &= \dfrac{s - b}{s - a}. \end{aligned}

However, D B D A = s b s a \frac{DB}{DA} = \frac{s - b}{s - a} as well. Since there only exists one point along line A B AB with this ratio, we conclude Z = D , Z = D, and D D lies on C P . CP. \, \blacksquare

We can finally find C P P D \frac{CP}{PD} , using Lemma 3, Lemma 4, and van Aubel's Theorem:

C P P D = C X X B + C Y Y A = s s a + s s b = 21 21 13 + 21 21 14 = 21 8 + 3 = 45 8 . \begin{aligned} \dfrac{CP}{PD} &= \dfrac{CX}{XB} + \dfrac{CY}{YA} \\ &= \dfrac{s}{s - a} + \dfrac{s}{s - b} \\ &= \dfrac{21}{21 - 13} + \dfrac{21}{21 - 14} \\ &= \dfrac{21}{8} + 3 \\ &= \dfrac{45}{8}. \end{aligned}

Our final answer is p + q = 45 + 8 = 53 . p + q = 45 + 8 = \boxed{53}.

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