Difference between revisions of "2008 AIME I Problems/Problem 10"

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Problem

Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$ . The diagonals have length $10\sqrt {21}$ , and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$ , respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$ . The distance $EF$ can be expressed in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ .

Solution

Solution 1

$[asy] size(300); defaultpen(1); pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0); draw(F--C--B--A); draw(E--A--D--C); draw(A--C,dashed); draw(circle(A,abs(C-A)),dotted); label("$A$",A,S); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE); label("$E$",E,N); label("$F$",F,S); clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); [/asy]$

Assuming that $ADE$ is a triangle and applying the triangle inequality, we see that $AD > 20\sqrt {7}$ . However, if $AD$ is strictly greater than $20\sqrt {7}$ , then the circle with radius $10\sqrt {21}$ and center $A$ does not touch $DC$ , which implies that $AC > 10\sqrt {21}$ , a contradiction. As a result, A, D, and E are collinear. Therefore, $AD = 20\sqrt {7}$ .

Thus, $ADC$ and $ACF$ are $30-60-90$ triangles. Hence $AF = 15\sqrt {7}$ , and

$EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}$

Finally, the answer is $25+7=\boxed{032}$ .

Solution 2

No restrictions are set on the lengths of the bases, so for calculational simplicity let $\angle CAF = 30^{\circ}$ . Since $CAF$ is a $30-60-90$ triangle, $AF=\frac{AC\sqrt{3}}2=15\sqrt{7}$ .

$EF=EA+AF=10\sqrt{7}+15\sqrt{7}=25\sqrt{7}$

The answer is $25+7=\boxed{032}$ . Note that while this is not rigorous, the above solution shows that $\angle CAF = 30^{\circ}$ is indeed the only possibility.

Solution 3

Extend $\overline {AB}$ through $B$ , to meet $\overline {DC}$ (extended through $C$ ) at $G$ . $ADG$ is an equilateral triangle because of the angle conditions on the base.

If $\overline {GC} = x$ then $\overline {CD} = 40\sqrt{7}-x$ , because $\overline{AD}$ and therefore $\overline{GD}$ $= 40\sqrt{7}$ .

By simple angle chasing, $CFD$ is a 30-60-90 triangle and thus $\overline{FD} = \frac{40\sqrt{7}-x}{2}$ , and $\overline{CF} = \frac{40\sqrt{21} - \sqrt{3}x}{2}$

Similarly $CAF$ is a 30-60-90 triangle and thus $\overline{CF} = \frac{10\sqrt{21}}{2} = 5\sqrt{21}$ .

Equating and solving for $x$ , $x = 30\sqrt{7}$ and thus $\overline{FD} = \frac{40\sqrt{7}-x}{2} = 5\sqrt{7}$ .

$\overline{ED}-\overline{FD} = \overline{EF}$

$30\sqrt{7} - 5\sqrt{7} = 25\sqrt{7}$ and $25 + 7 = \boxed{032}$

@@ Line 28: / Line 28: @@
 === Solution 2 ===
-No restrictions are set on the lengths of the bases, so for calculational simplicity let <math>\angle CAF = 30^{\circ}</math>. Since <math>CAF</math> is a <math>30-60-90</math> triangle, <math>AF=\frac{CF\sqrt{3}}2=15\sqrt{7}</math>.
+No restrictions are set on the lengths of the bases, so for calculational simplicity let <math>\angle CAF = 30^{\circ}</math>. Since <math>CAF</math> is a <math>30-60-90</math> triangle, <math>AF=\frac{AC\sqrt{3}}2=15\sqrt{7}</math>.
 <center><math>EF=EA+AF=10\sqrt{7}+15\sqrt{7}=25\sqrt{7}</math></center>
 The answer is <math>25+7=\boxed{032}</math>. Note that while this is not rigorous, the above solution shows that <math>\angle CAF = 30^{\circ}</math> is indeed the only possibility.

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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