Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 18: Line 18:
 
</asy>
 
</asy>
  
−
By the Power of a Point Theorem on <math>B</math>, we have <math>BD*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>
+
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_Geometry_AIME_2011_Problems/Problem_3&oldid=184555"