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

(Solution 1)
(Solution 1)
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Assuming that <math>ADE</math> is a triangle and applying the [[triangle inequality]], we see that <math>AD > 20\sqrt {7}</math>. However, if <math>AD</math> is strictly greater than <math>20\sqrt {7}</math>, then the circle with radius <math>10\sqrt {21}</math> and center <math>A</math> does not touch <math>DC</math>, which implies that <math>AC > 10\sqrt {21}</math>, a contradiction. As a result, A, D, and E are collinear. Therefore, <math>AD = 20\sqrt {7}</math>.
 
Assuming that <math>ADE</math> is a triangle and applying the [[triangle inequality]], we see that <math>AD > 20\sqrt {7}</math>. However, if <math>AD</math> is strictly greater than <math>20\sqrt {7}</math>, then the circle with radius <math>10\sqrt {21}</math> and center <math>A</math> does not touch <math>DC</math>, which implies that <math>AC > 10\sqrt {21}</math>, a contradiction. As a result, A, D, and E are collinear. Therefore, <math>AD = 20\sqrt {7}</math>.
  
It follows that <math>A</math>, <math>D</math>, and <math>E</math> are collinear, and also that <math>ADC</math> and <math>ACF</math> are <math>30-60-90</math> triangles. Hence <math>AF = 15\sqrt {7}</math>, and
+
Thus, <math>ADC</math> and <math>ACF</math> are <math>30-60-90</math> triangles. Hence <math>AF = 15\sqrt {7}</math>, and
 
<center><math>EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</math></center>
 
<center><math>EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</math></center>
 
Finally, the answer is <math>25+7=\boxed{032}</math>.
 
Finally, the answer is <math>25+7=\boxed{032}</math>.

Revision as of 17:41, 1 March 2009

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, $AF=\frac{CF\sqrt{3}}2=15\sqrt{7}$.

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

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.

See also

2008 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions