Difference between revisions of "2017 AMC 12B Problems/Problem 24"
m (→Solution) |
(→Solution) |
||
Line 15: | Line 15: | ||
<cmath>x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5</cmath> | <cmath>x^4-18x^2+1=0 \implies x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5</cmath> | ||
Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math> | Therefore, the answer is <math>\boxed{\textbf{(D) } 2+\sqrt{5}}</math> | ||
+ | |||
+ | |||
+ | [SOLUTION 2] Draw line FG through E, with F on BC and G on AD, FG//AB. WOLG let CD=1, CB=x, AB=x^2. By weighted average FG=(1+x^4)/(1+x^2). | ||
+ | |||
+ | Meanwhile, FE:EG=[CBE]:[ADE]=1:17. | ||
+ | FE=x^2/(1+x^2). We obtain (1+x^4)/(1+x^2)=18x^2/(1+x^2), | ||
+ | Namely x^4-18x^2+1=0. | ||
+ | |||
+ | The rest is the same with solution 1. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:48, 17 February 2017
Problem
Quadrilateral has right angles at and , is similar to , and . There exists a point in the interior of such that is similar to and the area of Triangle is times the area of Triangle . Find .
Solution
Let , , and . Note that . By the Pythagorean Theorem, . Since , the ratios of side lengths must be equal. Since , and . Let F be a point on such that is an altitude of triangle . Note that . Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields: Therefore, the answer is
[SOLUTION 2] Draw line FG through E, with F on BC and G on AD, FG//AB. WOLG let CD=1, CB=x, AB=x^2. By weighted average FG=(1+x^4)/(1+x^2).
Meanwhile, FE:EG=[CBE]:[ADE]=1:17. FE=x^2/(1+x^2). We obtain (1+x^4)/(1+x^2)=18x^2/(1+x^2), Namely x^4-18x^2+1=0.
The rest is the same with solution 1.
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.