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 "2009 AIME II Problems/Problem 3"
(→‎Solution)
Line 1: Line 1:
 
==Solution==
 
==Solution==
 
<center><asy>
 
<center><asy>
−
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10);
+
pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5);
 
draw (A--B--C--D--cycle);
 
draw (A--B--C--D--cycle);
 
pair E=(7,10);
 
pair E=(7,10);
Line 13: Line 13:
 
label("\(D\)",D,NE);
 
label("\(D\)",D,NE);
 
label("\(F\)",F,W);
 
label("\(F\)",F,W);
 +
label("\(100\)",Q,W);
 
</asy></center>
 
</asy></center>
−
From the problem, <math>AB=100</math> and triangle <math>AFB</math> is a right triangle. As <math>ABCD</math> is a rectangle, triangles <math>ADB</math>, and <math>ABE</math> are also right triangles. By <math>AA</math>, <math>\triangle AFB \sim \triangle ADB</math>, and by the altitude-on-hypotenuse theorems, <math>\triangle AFB \sim \triangle ABE</math>, so <math>\triangle ABE \sim \triangle ADB</math>. This gives <math>\frac {AE}{AB}= \frac {AB}{BC}</math>. <math>AE=\frac{AD}{2}</math> and <math>BD=AD</math>, so <math>\frac {AD}{2AB}= \frac {AB}{AD}</math>, or <math>(AD)^2=2(AB)^2</math>, so <math>AD=AB \sqrt{2}</math>, or <math>100 \sqrt{2}</math>, so the answer is <math>\boxed{141}</math>.
+
From the problem, <math>AB=100</math> and triangle <math>FBA</math> is a right triangle. As <math>ABCD</math> is a rectangle, triangles <math>BCA</math>, and <math>ABE</math> are also right triangles. By <math>AA</math>, <math>\triangle FBA \sim \triangle BCA</math>, and <math>\triangle FBA \sim \triangle ABE</math>, so <math>\triangle ABE \sim \triangle BCA</math>. This gives <math>\frac {AE}{AB}= \frac {AB}{BC}</math>. <math>AE=\frac{AD}{2}</math> and <math>BD=AD</math>, so <math>\frac {AD}{2AB}= \frac {AB}{AD}</math>, or <math>(AD)^2=2(AB)^2</math>, so <math>AD=AB \sqrt{2}</math>, or <math>100 \sqrt{2}</math>, so the answer is <math>\boxed{141}</math>.

Revision as of 20:15, 11 April 2009

Solution

[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label("\(E\)",E,N); label("\(A\)",A,NW); label("\(B\)",B,SW); label("\(C\)",C,SE); label("\(D\)",D,NE); label("\(F\)",F,W); label("\(100\)",Q,W); [/asy]

From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\triangle FBA \sim \triangle BCA$, and $\triangle FBA \sim \triangle ABE$, so $\triangle ABE \sim \triangle BCA$. This gives $\frac {AE}{AB}= \frac {AB}{BC}$. $AE=\frac{AD}{2}$ and $BD=AD$, so $\frac {AD}{2AB}= \frac {AB}{AD}$, or $(AD)^2=2(AB)^2$, so $AD=AB \sqrt{2}$, or $100 \sqrt{2}$, so the answer is $\boxed{141}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_3&oldid=31230"