Difference between revisions of "2008 AIME I Problems/Problem 10"
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== Problem == | == Problem == | ||
Let <math>ABCD</math> be an [[isosceles trapezoid]] with <math>\overline{AD}||\overline{BC}</math> whose angle at the longer base <math>\overline{AD}</math> is <math>\dfrac{\pi}{3}</math>. The [[diagonal]]s have length <math>10\sqrt {21}</math>, and point <math>E</math> is at distances <math>10\sqrt {7}</math> and <math>30\sqrt {7}</math> from vertices <math>A</math> and <math>D</math>, respectively. Let <math>F</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AD}</math>. The distance <math>EF</math> can be expressed in the form <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>. | Let <math>ABCD</math> be an [[isosceles trapezoid]] with <math>\overline{AD}||\overline{BC}</math> whose angle at the longer base <math>\overline{AD}</math> is <math>\dfrac{\pi}{3}</math>. The [[diagonal]]s have length <math>10\sqrt {21}</math>, and point <math>E</math> is at distances <math>10\sqrt {7}</math> and <math>30\sqrt {7}</math> from vertices <math>A</math> and <math>D</math>, respectively. Let <math>F</math> be the foot of the [[altitude]] from <math>C</math> to <math>\overline{AD}</math>. The distance <math>EF</math> can be expressed in the form <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>. | ||
− | == Solution == | + | == Solution 1 == |
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− | + | '''Key observation.''' <math>AD = 20\sqrt{7}</math>. | |
− | < | + | |
− | + | '''Proof 1.''' By the [[triangle inequality]], we can immediately see that <math>AD \geq 20\sqrt{7}</math>. However, notice that <math>10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}</math>, so by the law of sines, when <math>AD = 20\sqrt{7}</math>, <math>\angle ACD</math> is right and the circle centered at <math>A</math> with radius <math>10\sqrt{21}</math>, which we will call <math>\omega</math>, is tangent to <math>\overline{CD}</math>. Thus, if <math>AD</math> were increased, <math>\overline{CD}</math> would have to be moved even farther outwards from <math>A</math> to maintain the angle of <math>\frac{\pi}{3}</math> and <math>\omega</math> could not touch it, a contradiction. | |
− | + | ||
− | < | + | '''Proof 2.''' Again, use the triangle inequality to obtain <math>AD \geq 20\sqrt{7}</math>. Let <math>x = AD</math> and <math>y = CD</math>. By the law of cosines on <math>\triangle ADC</math>, <math>2100 = x^2+y^2-xy \iff y^2-xy+(x^2-2100) = 0</math>. Viewing this as a quadratic in <math>y</math>, the discriminant <math>\Delta</math> must satisfy <math>\Delta = x^2-4(x^2-2100) = 8400-3x^2 \geq 0 \iff x \leq 20\sqrt{7}</math>. Combining these two inequalities yields the desired conclusion. |
− | AD = 20\sqrt {7} | + | |
− | + | This observation tells us that <math>E</math>, <math>A</math>, and <math>D</math> are collinear, in that order. | |
− | < | + | |
− | EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</ | + | Then, <math>\triangle ADC</math> and <math>\triangle 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> | |
+ | Finally, the answer is <math>25+7=\boxed{032}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Extend <math>\overline {AB}</math> through <math>B</math>, to meet <math>\overline {DC}</math> (extended through <math>C</math>) at <math>G</math>. <math>ADG</math> is an equilateral triangle because of the angle conditions on the base. | ||
+ | |||
+ | If <math>\overline {GC} = x</math> then <math>\overline {CD} = 40\sqrt{7}-x</math>, because <math>\overline{AD}</math> and therefore <math>\overline{GD}</math> <math>= 40\sqrt{7}</math>. | ||
+ | |||
+ | By simple angle chasing, <math>CFD</math> is a 30-60-90 triangle and thus <math>\overline{FD} = \frac{40\sqrt{7}-x}{2}</math>, | ||
+ | and <math>\overline{CF} = \frac{40\sqrt{21} - \sqrt{3}x}{2}</math> | ||
+ | |||
+ | Similarly <math>CAF</math> is a 30-60-90 triangle and thus <math>\overline{CF} = \frac{10\sqrt{21}}{2} = 5\sqrt{21}</math>. | ||
+ | |||
+ | Equating and solving for <math>x</math>, <math>x = 30\sqrt{7}</math> and thus <math>\overline{FD} = \frac{40\sqrt{7}-x}{2} = 5\sqrt{7}</math>. | ||
+ | |||
+ | <math>\overline{ED}-\overline{FD} = \overline{EF}</math> | ||
+ | |||
+ | <math>30\sqrt{7} - 5\sqrt{7} = 25\sqrt{7}</math> and <math> 25 + 7 = \boxed{032}</math> | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:38, 28 November 2023
Contents
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 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)
Key observation. .
Proof 1. By the triangle inequality, we can immediately see that . However, notice that
, so by the law of sines, when
,
is right and the circle centered at
with radius
, which we will call
, is tangent to
. Thus, if
were increased,
would have to be moved even farther outwards from
to maintain the angle of
and
could not touch it, a contradiction.
Proof 2. Again, use the triangle inequality to obtain . Let
and
. By the law of cosines on
,
. Viewing this as a quadratic in
, the discriminant
must satisfy
. Combining these two inequalities yields the desired conclusion.
This observation tells us that ,
, and
are collinear, in that order.
Then, 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.