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Difference between revisions of "2009 AIME II Problems/Problem 3"

(New page: Image:AIMEII3.JPG From the problem, AB=100 and triangle AFB is a right triangle. As ABCD is a rectangle, triangles ADB, and ABE are also right triangles. By AA, AFB~ADB, and by the alt...)
 
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[[Image:AIMEII3.JPG]]
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==Solution==
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From the problem, AB=100 and triangle AFB is a right triangle. As ABCD is a rectangle, triangles ADB, and ABE are also right triangles. By AA, AFB~ADB, and by the altitude on hypotenuse theorems, AFB~ABE, so ABE~ADB. This gives AE/AB=AB/BC. AE=AD/2 and BD=AD, so AD/(2AB)=AB/AD, or (AD)^2=2(AB)^2, so AD=AB*sqrt(2), or 100*sqrt(2), so the answer is 141.
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<center><asy>
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pair A=(0,10), B=(0,0), C=(14,0), D=(14,10);
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draw (A--B--C--D--cycle);
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pair E=(7,10);
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draw (B--E);
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draw (A--C);
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pair F=(6.7,6.7);
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label("\(E\)",E,N);
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label("\(A\)",A,NW);
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label("\(B\)",B,SW);
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label("\(C\)",C,SE);
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label("\(D\)",D,NE);
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label("\(F\)",F,W);
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</asy></center>
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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>.

Revision as of 20:02, 11 April 2009

Solution

[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10); 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); [/asy]

From the problem, $AB=100$ and triangle $AFB$ is a right triangle. As $ABCD$ is a rectangle, triangles $ADB$, and $ABE$ are also right triangles. By $AA$, $\triangle AFB \sim \triangle ADB$, and by the altitude-on-hypotenuse theorems, $\triangle AFB \sim \triangle ABE$, so $\triangle ABE \sim \triangle ADB$. 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 $\boxed{141}$.

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