Difference between revisions of "2008 AIME I Problems/Problem 10"
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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 | + | Assuming that <math>ADE</math> is a triangle and applying the [[triangle inequality]], we see that <math>AD > 20\sqrt {7}</math>. The closest distance from the circle with radius <math>10\sqrt {21}</math> and center <math>A</math> to <math>\overline{CD}</math> is obtained by drawing a perpendicular to <math>\overline{CD}</math>. This creates a 30-60-90 triangle with base <math>20\sqrt{7}</math>. However, if <math>AD</math> is strictly greater than <math>20\sqrt {7}</math>, then the closest distance from <math>A</math> to <math>\overline{CD}</math> will be greater than <math>10\sqrt{21}</math>, a contradiction, because <math>AC = 10\sqrt{21}</math>. As a result, <math>AD = 20\sqrt {7}</math>, and A, D, and E are not a triangle at all, but are collinear. |
Thus, <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 |
Revision as of 19:33, 5 March 2013
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
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]](http://latex.artofproblemsolving.com/9/0/9/9091a509366e2723a7fb556a8255537bb23c13b1.png)
Assuming that is a triangle and applying the triangle inequality, we see that
. The closest distance from the circle with radius
and center
to
is obtained by drawing a perpendicular to
. This creates a 30-60-90 triangle with base
. However, if
is strictly greater than
, then the closest distance from
to
will be greater than
, a contradiction, because
. As a result,
, and A, D, and E are not a triangle at all, but are collinear.
Thus, and
are
triangles. Hence
, and
![$EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}$](http://latex.artofproblemsolving.com/d/2/3/d23856decb7cf64df12003938a5fb775faab773e.png)
Finally, the answer is .
Solution 2
No restrictions are set on the lengths of the bases, so for calculational simplicity let . Since
is a
triangle,
.
![$EF=EA+AF=10\sqrt{7}+15\sqrt{7}=25\sqrt{7}$](http://latex.artofproblemsolving.com/d/a/2/da2df7eee61c802dd5f20b93dd7cf26f70702634.png)
The answer is . Note that while this is not rigorous, the above solution shows that
is indeed the only possibility.
Solution 3
Extend through
, to meet
(extended through
) at
.
is an equilateral triangle because of the angle conditions on the base.
If then
, because
and therefore
.
By simple angle chasing, is a 30-60-90 triangle and thus
,
and
Similarly is a 30-60-90 triangle and thus
.
Equating and solving for ,
and thus
.
and
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 |