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Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 3"

(Created page with "==Problem== In triangle <math>ABC,</math> <math>BC=9.</math> Points <math>P</math> and <math>Q</math> are located on <math>BC</math> such that <math>BP=PQ=2,</math> <math>QC=5.<...")
 
 
(4 intermediate revisions by 2 users not shown)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
 +
<asy>
 +
unitsize(1cm);
 +
draw((-1,0)--(8,0)--(0,4)--cycle);
 +
draw((0,4)--(1,0));
 +
draw((0,4)--(3,0));
 +
label("$A$",(0,4),N);
 +
label("$B$",(-1,0),WSW);
 +
label("$C$",(8,0),ESE);
 +
label("$P$",(1,0),S);
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label("$Q$",(3,0),S);
 +
draw(circumcircle((0,4),(1,0),(3,0)));
 +
label("$D$",(-0.6,2),SW);
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label("$E$",(4.5,1.7),NE);
 +
</asy>
  
−
By the Power of a Point Theorem on <math>B</math>, we have <math>BD*BA=BP*BQ=2*4=8</math>. By the Power of a Point on <math>C</math>, we have <math>CE*CA=CQ*CP=5*7=35</math>. Dividing these two results yields <math>\frac{BD*BA}{CE*CA}=\frac{8}{35}</math>. We are given <math>BD=CE</math> and so <math>\frac{BD}{CE}=1</math>. Then the previous equation simplifies to <math>\frac{AB}{AC}=\frac{8}{35}</math>. Hence <math>m+n=8+35=\boxed{43}</math>
+
By the Power of a Point Theorem on <math>B</math>, we have <math>BD \times BA=BP \times BQ=2 \times 4=8</math>. By the Power of a Point on <math>C</math>, we have <math>CE \times CA=CQ \times CP=5 \times 7=35</math>. Dividing these two results yields <math>\frac{BD \times BA}{CE \times CA}=\frac{8}{35}</math>. We are given <math>BD=CE</math> and so <math>\frac{BD}{CE}=1</math>. Then the previous equation simplifies to <math>\frac{AB}{AC}=\frac{8}{35}</math>. Hence <math>m+n=8+35=\boxed{043}</math>
 +
 
 +
==Video Solution by Punxsutawney Phil==
 +
https://youtube.com/watch?v=OEkt8HbUVbQ&t=737s

Latest revision as of 21:05, 23 December 2022

Problem

In triangle $ABC,$ $BC=9.$ Points $P$ and $Q$ are located on $BC$ such that $BP=PQ=2,$ $QC=5.$ The circumcircle of $APQ$ cuts $AB,AC$ at $D,E$ respectively. If $BD=CE,$ then the ratio $\frac{AB}{AC}$ can be expressed in the form $\frac{m}{n},$ where $m,n$ are relatively prime positive integers. Find $m+n.$

Solution

[asy] unitsize(1cm); draw((-1,0)--(8,0)--(0,4)--cycle); draw((0,4)--(1,0)); draw((0,4)--(3,0)); label("$A$",(0,4),N); label("$B$",(-1,0),WSW); label("$C$",(8,0),ESE); label("$P$",(1,0),S); label("$Q$",(3,0),S); draw(circumcircle((0,4),(1,0),(3,0))); label("$D$",(-0.6,2),SW); label("$E$",(4.5,1.7),NE); [/asy]

By the Power of a Point Theorem on $B$, we have $BD \times BA=BP \times BQ=2 \times 4=8$. By the Power of a Point on $C$, we have $CE \times CA=CQ \times CP=5 \times 7=35$. Dividing these two results yields $\frac{BD \times BA}{CE \times CA}=\frac{8}{35}$. We are given $BD=CE$ and so $\frac{BD}{CE}=1$. Then the previous equation simplifies to $\frac{AB}{AC}=\frac{8}{35}$. Hence $m+n=8+35=\boxed{043}$

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=OEkt8HbUVbQ&t=737s

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