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In triangle $ABC$, $P$ and $Q$ are points on $AB$ and $AC$ respectively such that $AP:PB=8:1$ and $AQ:QC=15:1$. $X$ and $Y$ are points on $BC$ such that the circumcircle of triangle $APX$ is tangent to both $BC$ and $CA$, while the circumcircle of triangle $AQY$ is tangent to both $AB$ and $BC$.

How to find $\cos\widehat{BAC}$?

enter image description here

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    $\begingroup$ Use math.meta.stackexchange.com/questions/5020/… to properly format equations. $\endgroup$ – Arbuja Apr 20 '17 at 14:00
  • $\begingroup$ What is Mathjax? How to use it? $\endgroup$ – Ray Cheng Apr 20 '17 at 14:00
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    $\begingroup$ @RayCheng: I have improved formatting (have a look at the code), but you still need to add you attempts. $\endgroup$ – Jack D'Aurizio Apr 20 '17 at 14:25
  • $\begingroup$ Thanks, @Jack D'Aurizio for this nice figure. I imagine as you have built point I because it contributes to a solution ? $\endgroup$ – Jean Marie Apr 20 '17 at 14:31
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    $\begingroup$ I am a new user. Thank you for your kind support! $\endgroup$ – Ray Cheng Apr 20 '17 at 14:53
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my solution again but in a neater way enter image description here

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Let $O_C$ be the center of the circumcircle of $APX$ and $O_B$ be the center of the circumcircle of $AQY$.

$O_C$ has to lie on three different lines: the perpendicular to $CA$ through $A$, the angle bisector of $\widehat{ACB}$ and the perpendicular bisector of $AP$. Similarly, $O_B$ has to lie on the perpendicular to $BA$ through $A$, on the angle bisector of $\widehat{CBA}$ and on the perpendicular bisector of $AQ$.

Since $O_C$ lies on a angle bisector and on a perpendicular, $O_C A$ is given by $b\tan\frac{C}{2}$.
Since $O_C$ lies on the perpendicular bisector of $AP$, we also have $O_C A\sin A=\frac{4}{9}c$.

A similar argument for $O_B$ leads to $O_B A = c\tan\frac{B}{2}$ and $O_B A\sin A=\frac{15}{32}b$. Summarizing:

$$\left\{\begin{array}{rcl}b \tan\frac{C}{2}&=&\frac{4c}{9\sin A}\\c\tan\frac{B}{2}&=&\frac{15b}{32\sin A} \end{array}\right. $$ that can be written as $$\left\{\begin{array}{rcl}\frac{br}{a+b-c}&=&\frac{4c R}{9a}\\ \frac{cr}{a+c-b}&=&\frac{15b R}{32 a} \end{array}\right. $$ implying $$ c^2=18\cdot\frac{\Delta^2}{(a+b)^2-c^2},\qquad b^2=\frac{256}{15}\cdot\frac{\Delta^2}{(a+c)^2-b^2} $$ and $$ c^2=\frac{9}{8}(a-b+c)(-a+b+c),\qquad b^2=\frac{16}{15}(-a+b+c)(a+b-c) $$ due to Heron's formula. This implies $$ (a,b,c)=\lambda\cdot (13,8,15) $$ and

$$ \cos\widehat{BAC} = \frac{b^2+c^2-a^2}{2bc} = \color{red}{\frac{1}{2}}. $$

$\widehat{B}$ is not a right angle but it is quite close to a right angle, since $8^2+13^2=15^2+8$.

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  • $\begingroup$ How to find b/c $\endgroup$ – Ray Cheng Apr 21 '17 at 0:36
  • $\begingroup$ also is angle B a right angle? $\endgroup$ – Ray Cheng Apr 21 '17 at 0:37
  • $\begingroup$ In order to find the b/c ratio you just to divide the last two equations term by term and solve a quadratic polynomial. $\endgroup$ – Jack D'Aurizio Apr 21 '17 at 5:47
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    $\begingroup$ a=2R sin A comes from Inscribed Angle Theorem proofwiki.org/wiki/Law_of_Sines $\endgroup$ – Ray Cheng Apr 21 '17 at 10:34
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    $\begingroup$ tan(C/2)=2r/(a+b-c) comes from the law of cotangents. en.wikipedia.org/wiki/Law_of_cotangents $\endgroup$ – Ray Cheng Apr 21 '17 at 10:35
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@Jack D'Aurizio: The correct answer should be 1/6 but not 1/2, and this is the solution given by The Hong Kong Academy for Gifted Education (HKAGE). enter image description here

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