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

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Triangle <math>ABC</math> has side lengths <math>AB=7, BC=8, </math> and <math>CA=9.</math> Circle <math>\omega_1</math> passes through <math>B</math> and is tangent to line <math>AC</math> at <math>A.</math> Circle <math>\omega_2</math> passes through <math>C</math> and is tangent to line <math>AB</math> at <math>A.</math> Let <math>K</math> be the intersection of circles <math>\omega_1</math> and <math>\omega_2</math> not equal to <math>A.</math> Then <math>AK=m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
 
Triangle <math>ABC</math> has side lengths <math>AB=7, BC=8, </math> and <math>CA=9.</math> Circle <math>\omega_1</math> passes through <math>B</math> and is tangent to line <math>AC</math> at <math>A.</math> Circle <math>\omega_2</math> passes through <math>C</math> and is tangent to line <math>AB</math> at <math>A.</math> Let <math>K</math> be the intersection of circles <math>\omega_1</math> and <math>\omega_2</math> not equal to <math>A.</math> Then <math>AK=m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
  
==Solution==
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==Solution 1==
 +
<asy>
 +
unitsize(20);
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pair B = (0,0);
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pair A = (2,sqrt(45));
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pair C = (8,0);
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draw(circumcircle(A,B,(-17/8,0)),rgb(.7,.7,.7));
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draw(circumcircle(A,C,(49/8,0)),rgb(.7,.7,.7));
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draw(B--A--C--cycle);
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label("$A$",A,dir(105));
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label("$B$",B,dir(-135));
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label("$C$",C,dir(-75));
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dot((2.68,2.25));
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label("$K$",(2.68,2.25),dir(-150));
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label("$\omega_1$",(-6,1));
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label("$\omega_2$",(14,6));
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label("$7$",(A+B)/2,dir(140));
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label("$8$",(B+C)/2,dir(-90));
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label("$9$",(A+C)/2,dir(60));
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</asy>
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-Diagram by Brendanb4321
 +
 
 +
 
 
Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath> Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}</math>. This gives us <cmath>AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49</cmath> <cmath>\implies \frac{196}{81}AK^2=49</cmath> <cmath>AK=\frac{9}{2}</cmath> so our answer is <math>9+2=\boxed{011}</math>.
 
Note that from the tangency condition that the supplement of <math>\angle CAB</math> with respects to lines <math>AB</math> and <math>AC</math> are equal to <math>\angle AKB</math> and <math>\angle AKC</math>, respectively, so from tangent-chord, <cmath>\angle AKC=\angle AKB=180^{\circ}-\angle BAC</cmath> Also note that <math>\angle ABK=\angle KAC</math>, so <math>\triangle AKB\sim \triangle CKA</math>. Using similarity ratios, we can easily find <cmath>AK^2=BK*KC</cmath> However, since <math>AB=7</math> and <math>CA=9</math>, we can use similarity ratios to get <cmath>BK=\frac{7}{9}AK, CK=\frac{9}{7}AK</cmath> Now we use Law of Cosines on <math>\triangle AKB</math>: From reverse Law of Cosines, <math>\cos{\angle BAC}=\frac{11}{21}\implies \cos{(180^{\circ}-\angle BAC)}=-\frac{11}{21}</math>. This gives us <cmath>AK^2+\frac{49}{81}AK^2+\frac{22}{27}AK^2=49</cmath> <cmath>\implies \frac{196}{81}AK^2=49</cmath> <cmath>AK=\frac{9}{2}</cmath> so our answer is <math>9+2=\boxed{011}</math>.
 
-franchester
 
-franchester

Revision as of 09:38, 23 March 2019

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 $AK=m/n,$ 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

See Also

2019 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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