AIME 2018 II · 第 14 题

Solution 1

Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$, respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$. Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$, it follows that $\angle AYP = \angle QYX = \angle YXC = \beta$. By equal tangents, $PZ = PY$. Applying the Law of Sines to $\triangle APY$ yields

$$\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.$$ Similarly, applying the Law of Sines to $\triangle ABX$ gives

$$\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.$$ It follows that

$$2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,$$ implying $AZ = \tfrac{21}5$. Applying the same argument to $\triangle AQY$ yields

$$2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),$$ from which $AQ = \tfrac{168}{59}$. The requested sum is $168 + 59 = \boxed{227}$.

Solution 2 (Projective)

Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $Z$ and $W$. By Brianchon's theorem on tangential hexagons $QWCBZP$ and $PYQCXB$, we know that $ZW,CP,BQ$ and $XY$ are concurrent at a point $O$. Let $PQ \cap BC = M$. Then by La Hire's $A$ lies on the polar of $M$ so $M$ lies on the polar of $A$. Therefore, $ZW$ also passes through $M$. Then projecting through $M$, we have

$$-1 = (A,O;Y,X) \stackrel{M}{=} (A,Z;P,B) \stackrel{M}{=} (A,W;Q,C).$$ Therefore, $\frac{AP \cdot ZB}{MP \cdot AB} = 1 \implies \frac{3 \cdot ZB}{ZP \cdot 7} = 1$. Since $ZB+ZP=4$ we know that $ZP = \frac{6}{5}$ and $ZB = \frac{14}{5}$. Therefore, $AW = AZ = \frac{21}{5}$ and $WC = 8 - \frac{21}{5} = \frac{19}{5}$. Since $(A,W;Q,C) = -1$, we also have $\frac{AQ \cdot WC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1$. Solving for $AQ$, we obtain $AQ = \frac{168}{59} \implies m+n = \boxed{227}$. 😃 -Vfire

Solution 3 (Combination of Law of Sine and Law of Cosine)

Let the center of the incircle of $\triangle ABC$ be $O$. Link $OY$ and $OX$. Then we have $\angle OYP=\angle OXB=90^{\circ}$

$\because$ $OY=OX$

$\therefore$ $\angle OYX=\angle OXY$

$\therefore$ $\angle PYX=\angle YXB$

$\therefore$ $\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA$

Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $M$ and $N$, let $MP=YP=x$ and $NQ=YQ=y$.

Use Law of Sine in $\triangle APY$ and $\triangle AXB$, we have

$\frac{\sin \angle PAY}{PY}=\frac{\sin \angle PYA}{PA}$ $\frac{\sin \angle BAX}{BX}=\frac{\sin \angle AXB}{AB}$

therefore we have

$\frac{3}{x}=\frac{7}{4-x}$

Solve this equation, we have $x=\frac{6}{5}$

As a result, $MB=4-x=\frac{14}{5}=BX$, $AM=x+3=\frac{21}{5}=AN$, $NC=8-AN=\frac{19}{5}=XC$, $AQ=\frac{21}{5}-y$, $PQ=\frac{6}{5}+y$

So, $BC=\frac{14}{5}+\frac{19}{5}=\frac{33}{5}$

Use Law of Cosine in $\triangle BAC$ and $\triangle PAQ$, we have

$\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}$ $\cos \angle PAQ=\frac{AP^2+AQ^2-PQ^2}{2\cdot AP\cdot AQ}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$

And we have

$\cos \angle BAC=\cos \angle PAQ$

So

$\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$

Solve this equation, we have $y=\frac{399}{295}=QN$

As a result, $AQ=AN-QN=\frac{21}{5}-\frac{399}{295}=\frac{168}{59}$

So, the final answer of this question is $168+59=\boxed {227}$

~Solution by $BladeRunnerAUG$ (Frank FYC)

Solution 4 (Projective geometry)

Claim

Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $M$ and $N$, respectively. Then lines $PQ, MN,$ and $BC$ are concurrent and lines $PC, MN, AX,$ and $BQ$ are concurrent.

Proof

Let $E$ be point of crossing $AX$ and $BQ.$ We make projective transformation such that circle $\omega$ maps into the circle and point $E$ maps into the center of new circle point $I.$ We denote images using notification $X \rightarrow X'.$

$BCQP$ maps into $B'C'Q'P'$, so lines $B'Q'$ and $A'X'$ be the diameters. This implies $P'Q'||B'C', \angle B'P'Q' = \angle B'C'Q' = 90^\circ \implies B'C'Q'P'$ be a square.

Therefore $M'N'$ be the diameter $\implies P'C', B'Q',$ be diagonals of the square. $M'N'$ and $X'Y'$ be midlines which crossing in the center $I.$ Therefore lines $PC, MN, AX,$ and $BQ$ are concurrent.

Lines $P'Q'||M'N' ||B'C' \implies PQ, MN$ and $BC$ are concurrent.

Solution The cross-ratio associated with a list of four collinear points $A,P,M,D$ is defined as

$$(A,P;M,B)={\frac {AP\cdot MB}{AB\cdot PM}}.$$ The cross-ratio be projective invariant of a quadruple of collinear points, so

$$(A,P; M,B) = {\frac {A'P'\cdot M'B'}{A'B'\cdot P'M'}} = \frac {M'B'}{P'M'} = 1.$$ $$(A,P; M,B)={\frac {3\cdot (7 - AM)}{7\cdot (AM - 3)}} = 1 \implies AM = \frac {21}{5} \implies AN = AM = \frac {21}{5}.$$

$$(A,Q;N,C)={\frac {AQ\cdot NC}{AC\cdot QN}} = \frac {AQ\cdot (AC- AN)}{AC\cdot (AN-AQ)} = 1.$$ $$AQ \cdot (8 - \frac{21}{5}) = 8 \cdot (\frac{21}{5} – AQ) \implies AQ = \frac{168}{59}.$$ For visuals only, I will show how one can find the perceptor $D$ and the image’s plane. $E_0$ is image of inversion $E$ with respect $\omega.$ $UW$ is the diameter of $\omega, E,E_0,U,W$ are collinear. $DU \perp \omega, DE_0\perp WD, UV \perp WD, UV$ is diameter of $\omega'$.

Plane of images is perpendicular to $WD.$

Last diagram shows the result of transformation. Transformation is possible. The end.

vladimir.shelomovskii@gmail.com, vvsss

Solution 5

Firstly, assume $PY=x=PZ, ZB=4-x=BX, AZ=AW=3+x, CW=CX=5-x, QY=QW=y, AQ=3+x-y$

By tangency, we have $\angle{PYZ}=\angle{PZY}=\angle{YXZ}; \angle{BZX}=\angle{BXZ}=\angle{ZYX}; \angle{PYX}=\angle{BXY}$

Similar reason yields $\angle{QYX}=\angle{CXY}$. Apply Law of sines

We have $\frac{3}{\sin{\angle{PYA}}}=\frac{x}{\sin{\angle{PAY}}}, \frac{7}{\sin{\angle{AXB}}}=\frac{4-x}{\sin{\angle{BAX}}}$ Since $\angle{PYA}=180-\angle{BXA}$ so their sine values would be the same. Solve this system and we have $\frac{3}{7}=\frac{x}{4-x}, x=\frac{6}{5}$

Apply the same process in $\triangle{AQY}, \triangle{AXC}$, we have $\frac{3+x-y}{8}=\frac{y}{5-x}, y=\frac{399}{295}$

The desired length is $3+x-y=\frac{168}{59}\implies \boxed{227}$

~Bluesoul

Solution 6(Algebra Bash)

Let $\angle PAC = \theta$. Let $PY = a$ and $YQ = c$. By PoP and Stewart's Theorem

$(4 - a)(9 - 2a)(5 - a) + AX^2 (9 - 2a) = 64(4 - a) + 49(5 - a)$ where we will let $AX = u$ for simplicity. Now let $AY = x$. By LoC we have

$9 + x^2 - 6x\cos(\theta) = a^2$

By LoC we also have

$49 + AX^2 - 14 AX \cos(\theta) = (4 - a)^2$

Finally, by PoP, we have

$x \cdot AX = (a + 3)^2$

Solving this system of equations, we get

$a = \frac{6}{5}, x = \frac{63 \cdot \sqrt{2761}}{1255}$. Note that by PoP, $AQ = a + 3 - c$ by using the Power of Point of tangents that are equal. Finally, by Stewart's Theorem on $\triangle APQ$ we have

$a(a + c)c + x^2 (a + c) = (a + 3 - c)^2 a + 9c$ in which after plugging in the values we found for $a$ and $x$ we get $c = \frac{399}{295}$. Finally, $a + 3 - c = \frac{168}{59} \implies \boxed{227}$.

~ilikemath247365

Video Solution by Mop 2024

https://youtu.be/SIs1JFLFzyw

~r00tsOfUnity