Triangle has The -excircle of is tangent to the extensions of and at and respectively, and is tangent to at Now, extend to meet at and, similarly, extend to meet at Letting be the point where and intersect, the ratio is equal to where and are coprime positive integers.
Find
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
(Geogebra couldn't let me input A ′ and 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 . Also, we will let B C = a , C A = b , A B = c , and s = 2 a + b + c .
First, we'll prove this lemma:
Lemma 1: A B A ′ B ′ = 2 cos 2 A cos 2 B .
Consider a general acute triangle A B C , and let the C -excircle with center E be tangent at A ′ , B ′ , D as shown in the diagram.
Let I be the incenter of △ A B C . Note that
∠ B ′ A ′ D = 2 1 ∠ B ′ E D = 2 1 ( 1 8 0 − ∠ D B B ′ ) = 2 B = ∠ I B A .
Similarly, ∠ A ′ B D = 2 A = ∠ I A B . Therefore, △ A ′ B ′ D ∼ △ B A I .
We can find that A ′ D = 2 ( s − a ) cos 2 A and B I = cos 2 B s − a , so
A B A ′ B ′ = B I A ′ D = cos 2 B s − a 2 ( s − a ) cos 2 A = 2 cos 2 A cos 2 B . ■
Next, we will find X C B X and Y C A Y . To accomplish this, we will find the lengths of B ′ X , X C , A ′ Y , Y C via the Ratio Lemma, then use the facts that B X = B ′ X − B ′ B and A Y = A ′ Y − A ′ A .
Before moving on, we note that ∠ C A ′ X = 2 A and ∠ C B ′ Y = 2 B .
Lemma 2: X C B ′ X = a − a + b + c , Y C A ′ Y = b a − b + c
By the results of Lemma 1, we have A ′ B ′ = 2 c cos 2 A cos 2 B . Now, applying the Ratio Lemma for X C B ′ X in △ A ′ B ′ C , we have
X C B ′ X = A ′ C A ′ B ′ ⋅ sin ∠ C A ′ X sin ∠ B ′ A ′ X = s 2 c cos 2 A cos 2 B ⋅ sin 2 A sin 2 B = s sin 2 A c cos 2 A sin B .
We know that cos 2 A = b c s ( s − a ) and sin 2 A = b c ( s − b ) ( s − c ) :
s sin 2 A c cos 2 A sin B = s c sin B ( s − b ) ( s − c ) s ( s − a ) = s ( s − a ) ( s − b ) ( s − c ) c ( s − a ) sin B .
We recognize the denominator as [ A B C ] , by Heron's formula. By another area formula, we also have sin B = a c 2 [ A B C ] . Thus,
X C B ′ X = s ( s − a ) ( s − b ) ( s − c ) c ( s − a ) sin B = [ A B C ] c ( s − a ) a c 2 [ A B C ] = a 2 ( s − a ) = a − a + b + c .
Using the same strategy, we can find Y C A ′ Y = b a − b + c . ■
Now, we can find X C B X and Y C A Y :
Lemma 3: X C B X = s s − a , Y C A Y = s s − b
Again, we'll focus solely on the points and segments on ray C B . By Lemma 2, B ′ X = a − a + b + c X C . We also have B ′ X + X C = s , by the properties of the excircle. Thus,
B ′ X + X C X C ( a − a + b + c + 1 ) X C ( a b + c ) X C B ′ X = s = s = s = b + c s a . = s − b + c s a .
Next, we apply B X = B ′ X − B ′ B :
B X = B ′ X − B ′ B = s − b + c s a − ( s − a ) = a − b + c s a = a ( 1 − 2 ( b + c ) a + b + c ) = 2 ( b + c ) a ( − a + b + c ) = b + c a ( s − a ) .
Finally,
X C B X = b + c s a b + c a ( s − a ) = s s − a .
In a similar vein, we can find that Y C A Y = s s − b . ■
Here's the last observation before we can calculate our final answer:
Lemma 4: C , P , D are collinear.
Suppose C P intersects A B at point Z . By Ceva's Theorem, Z A Z B ⋅ X C X B ⋅ Y A Y C = 1 . Applying the results of Lemma 3 yields
Z A Z B ⋅ X C X B ⋅ Y A Y C Z A Z B ⋅ s s − a ⋅ s − b s Z A Z B = 1 = 1 = s − a s − b .
However, D A D B = s − a s − b as well. Since there only exists one point along line A B with this ratio, we conclude Z = D , and D lies on C P . ■
We can finally find P D C P , using Lemma 3, Lemma 4, and van Aubel's Theorem:
P D C P = X B C X + Y A C Y = s − a s + s − b s = 2 1 − 1 3 2 1 + 2 1 − 1 4 2 1 = 8 2 1 + 3 = 8 4 5 .
Our final answer is p + q = 4 5 + 8 = 5 3 .