Difference between revisions of "2005 AIME I Problems/Problem 7"
(2 solutions, box) |
|||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | In quadrilateral <math> ABCD, BC=8, CD=12, AD=10, </math> and <math> m\angle A= m\angle B = 60^\circ. </math> Given that <math> AB = p + \sqrt{q}, </math> where <math> p </math> and <math> q </math> are positive | + | In [[quadrilateral]] <math> ABCD,\ BC=8,\ CD=12,\ AD=10, </math> and <math> m\angle A= m\angle B = 60^\circ. </math> Given that <math> AB = p + \sqrt{q}, </math> where <math> p </math> and <math> q </math> are [[positive]] [[integer]]s, find <math> p+q. </math> |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
+ | {{image}} | ||
+ | Draw the [[perpendicular]]s from <math>C</math> and <math>D</math> to <math>AB</math>, labeling the intersection points as <math>E</math> and <math>F</math>. This forms 2 <math>30-60-90</math> [[right triangle]]s, so <math>AE = 5</math> and <math>BF = 4</math>. Also, if we draw the horizontal line extending from <math>C</math> to a point <math>G</math> on the line <math>DE</math>, we find another right triangle. <math>DG = DE - CF = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3}</math>. The [[Pythagorean theorem]] yields that <math>GC^2 = 12^2 - \sqrt{3}^2 = 141</math>, so <math>EF = GC = \sqrt{141}</math>. Therefore, <math>AB = 5 + 4 + \sqrt{141} = 9 + \sqrt{141}</math>, and <math>p + q = 150</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | {{image}} | ||
+ | Extend <math>AD</math> and <math>BC</math> to an intersection at point <math>E</math>. We get an [[equilateral triangle]] <math>ABE</math>. Solve <math>\triangle CDP</math> using the [[Law of Cosines]], denoting the length of a side of <math>\triangle ABE</math> as <math>s</math>. We get <math>12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\cos 60 \Longrightarrow 144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)</math>. This boils down a [[quadratic equation]]: <math>0 = s^2 - 18s + 60</math>; the [[quadratic formula]] yields the (discard the negative result) same result of <math>9 + \sqrt{141}</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2005|n=I|num-b=6|num-a=8}} | |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 15:49, 4 March 2007
Problem
In quadrilateral and
Given that
where
and
are positive integers, find
Solution
Solution 1
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
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 2
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Extend and
to an intersection at point
. We get an equilateral triangle
. Solve
using the Law of Cosines, denoting the length of a side of
as
. We get
. This boils down a quadratic equation:
; the quadratic formula yields the (discard the negative result) same result 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 |