Difference between revisions of "2019 AIME II Problems/Problem 11"

Revision as of 17:53, 7 August 2020

Problem

Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

$[asy] unitsize(20); pair B = (0,0); pair A = (2,sqrt(45)); pair C = (8,0); draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7)); draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7)); draw(B--A--C--cycle); label("$A$",A,dir(105)); label("$B$",B,dir(-135)); label("$C$",C,dir(-75)); dot((2.68,2.25)); label("$K$",(2.68,2.25),dir(-150)); label("$\omega_1$",(-6,1)); label("$\omega_2$",(14,6)); label("$7$",(A+B)/2,dir(140)); label("$8$",(B+C)/2,dir(-90)); label("$9$",(A+C)/2,dir(60)); [/asy]$ -Diagram by Brendanb4321

Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$ , respectively, so from tangent-chord, $\angle AKC=\angle AKB=180^{\circ}-\angle BAC$ Also note that $\angle ABK=\angle KAC$ , so $\triangle AKB\sim \triangle CKA$ . Using similarity ratios, we can easily find $AK^2=BK*KC$ However, since $AB=7$ and $CA=9$ , we can use similarity ratios to get $BK=\frac{7}{9}AK, CK=\frac{9}{7}AK$ Now we use Law of Cosines on $\triangle AKB$ : From reverse Law of Cosines, $\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}$ . This gives us $AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49$ $\implies \frac{196}{81}AK^2=49$ $AK=\frac{9}{2}$ so our answer is $9+2=\boxed{011}$ . -franchester

Solution 2 (Inversion)

Consider an inversion with center $A$ and radius $r=AK$ . Then, we have $AB\cdot AB^*=AK^2$ , or $AB^*=\frac{AK^2}{7}$ . Similarly, $AC^*=\frac{AK^2}{9}$ . Notice that $AB^*KC^*$ is a parallelogram, since $\omega_1$ and $\omega_2$ are tangent to $AC$ and $AB$ , respectively. Thus, $AC^*=B^*K$ . Now, we get that $\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}$ so by Law of Cosines on $\triangle AB^*K$ we have $(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)$ $\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}$ $\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}$ $\Rightarrow AK=\frac{9}{2}$ Then, our answer is $9+2=\boxed{11}$ . -brianzjk

Solution 3 (Video)

Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI

@@ Line 28: / Line 28: @@
 -franchester
-==Solution 2 (Video)==
+==Solution 2 (Inversion)==
+Consider an inversion with center <math>A</math> and radius <math>r=AK</math>. Then, we have <math>AB\cdot AB^*=AK^2</math>, or <math>AB^*=\frac{AK^2}{7}</math>. Similarly, <math>AC^*=\frac{AK^2}{9}</math>. Notice that <math>AB^*KC^*</math> is a parallelogram, since <math>\omega_1</math> and <math>\omega_2</math> are tangent to <math>AC</math> and <math>AB</math>, respectively. Thus, <math>AC^*=B^*K</math>. Now, we get that
+<cmath>\cos(\angle AB^*K)=-\cos(180-\angle BAC)=-\frac{11}{21}</cmath>
+so by Law of Cosines on <math>\triangle AB^*K</math> we have
+<cmath>(AK)^2=(AB^*)2+(B^*K)^2-2\cdot AB^*\cdot B^*K \cdot \cos(\angle AB^*K)</cmath>
+<cmath>\Rightarrow (AK)^2=\frac{AK^4}{49}+\frac{AK^4}{81}-2\cdot \frac{AK^2}{7}\frac{AK^2}{9}\frac{-11}{21}</cmath>
+<cmath>\Rightarrow 1=\frac{AK^2}{49}+\frac{AK^2}{81}+\frac{11AK^2}{63\cdot21}</cmath>
+<cmath>\Rightarrow AK=\frac{9}{2}</cmath>
+Then, our answer is <math>9+2=\boxed{11}</math>.
+-brianzjk
+==Solution 3 (Video)==
 Video Link: https://www.youtube.com/watch?v=nJydO5CLuuI

2019 AIME II (Problems • Answer Key • Resources)
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