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 ( | + | 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>. |
=== Solution 2 === | === Solution 2 === |
Latest revision as of 19:20, 3 March 2020
Problem
In quadrilateral and Given that where and are positive integers, find
Contents
[hide]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
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
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 |
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