Difference between revisions of "2008 AIME I Problems/Problem 10"
(solution by Boy Soprano II) |
m (→Solution: correct asy) |
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Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
<center><asy> | <center><asy> | ||
− | size( | + | size(300); |
defaultpen(1); | 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); | 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); | ||
Line 11: | Line 11: | ||
draw(A--C,dashed); | draw(A--C,dashed); | ||
draw(circle(A,abs(C-A)),dotted); | draw(circle(A,abs(C-A)),dotted); | ||
− | label(" | + | label("\(A\)",A,S); |
− | label(" | + | label("\(B\)",B,NW); |
− | label(" | + | label("\(C\)",C,NE); |
− | label(" | + | label("\(D\)",D,SE); |
− | label(" | + | label("\(E\)",E,W); |
− | label(" | + | label("\(F\)",F,S); |
clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); | clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); | ||
</asy></center> | </asy></center> | ||
Line 26: | Line 26: | ||
<cmath> | <cmath> | ||
AD = 20\sqrt {7}.</cmath> | AD = 20\sqrt {7}.</cmath> | ||
− | It follows that <math>A,D,E</math> are collinear, and also that <math>ADC</math> and <math>ACF</math> are 30-60-90 triangles. Hence <math>AF = 15\sqrt {7}</math>, and | + | It follows that <math>A,D,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 |
<cmath> | <cmath> | ||
EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</cmath> | EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</cmath> | ||
− | Hence the answer to this problem is 25+7, or | + | Hence the answer to this problem is <math>25+7</math>, or $\boxed{032}. |
== See also == | == See also == |
Revision as of 13:58, 23 March 2008
Problem
Let be an isosceles trapezoid with
whose angle at the longer base
is
. The diagonals have length
, and point
is at distances
and
from vertices
and
, respectively. Let
be the foot of the altitude from
to
. The distance
can be expressed in the form
, where
and
are positive integers and
is not divisible by the square of any prime. Find
.
Solution
![[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,W); label("\(F\)",F,S); clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); [/asy]](http://latex.artofproblemsolving.com/8/5/b/85b1047106c06f1ef0e539451f9824fe436b26bd.png)
Applying the triangle inequality to , we see that
On the other hand, if
is strictly greater than
, then the circle with radius
and center
does not touch
, which implies that
, a contradiction. Hence
It follows that
are collinear, and also that
and
are
triangles. Hence
, and
Hence the answer to this problem is
, or $\boxed{032}.
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
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 |