Difference between revisions of "2005 AIME I Problems/Problem 7"
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+ | Draw line segment <math>DE</math> such that line <math>DE</math> is concurrent with line <math>BC</math>. Then, <math>ABED</math> is an isosceles trapezoid so <math>AD=BE=10</math>, and <math>BC=8</math> and <math>EC=2</math>. We are given that <math>DC=12</math>. Since <math>\angle CED = 120^{\circ}</math>, using Law of Cosines on <math>\bigtriangleup CED</math> gives <cmath>12^2=DE^2+4-2(2)(DE)(\cos 120^{\circ})</cmath> which gives <cmath>144-4=DE^2+2DE</cmath>. Adding <math>1</math> to both sides gives <math>141=(DE+1)^2</math>, so <math>DE=\sqrt{141}-1</math>. <math>\bigtriangleup DAP</math> and <math>\bigtriangleup EBQ</math> are both <math>30-60-90</math>, so <math>AP=5</math> and <math>BQ=5</math>. <math>PQ=DE</math>, and therefore <math>AB=AP+PQ+BQ=5+\sqrt{141}-1+5=9+\sqrt{141} \rightarrow (p,q)=(9,141) \rightarrow \boxed{150}</math>. | ||
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=== Solution 2 === | === Solution 2 === | ||
<center>[[Image:AIME_2005I_Solution_7_1.png]]</center> | <center>[[Image:AIME_2005I_Solution_7_1.png]]</center> |
Latest revision as of 20:20, 3 March 2020
Problem
In quadrilateral and
Given that
where
and
are positive integers, find
Contents
Solution
Solution 1
Draw line segment such that line
is concurrent with line
. Then,
is an isosceles trapezoid so
, and
and
. We are given that
. Since
, using Law of Cosines on
gives
which gives
. Adding
to both sides gives
, so
.
and
are both
, so
and
.
, and therefore
.
Solution 2
![AIME 2005I Solution 7 1.png](https://wiki-images.artofproblemsolving.com//4/41/AIME_2005I_Solution_7_1.png)
Draw the perpendiculars from and
to
, labeling the intersection points as
and
. This forms 2
right triangles, so
and
. Also, if we draw the horizontal line extending from
to a point
on the line
, we find another right triangle
.
. The Pythagorean Theorem yields that
, so
. Therefore,
, and
.
Solution 3
![AIME 2005I Solution 7 2.png](https://wiki-images.artofproblemsolving.com//2/2f/AIME_2005I_Solution_7_2.png)
Extend and
to an intersection at point
. We get an equilateral triangle
. We denote the length of a side of
as
and solve for it using the Law of Cosines:
This simplifies to
; the quadratic formula yields the (discard the negative result) same result of
.
Solution 4
Extend and
to meet at point
, forming an equilateral triangle
. Draw a line from
parallel to
so that it intersects
at point
. Then, apply Stewart's Theorem on
. Let
.
By the quadratic formula (discarding the negative result),
, giving
for a final answer of
.
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.