Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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==Solution== | ==Solution== | ||
− | Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD | + | Let <math>CD=1</math>, <math>BC=x</math>, and <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. By the Pythagorean Theorem, <math>BD=\sqrt{x^2+1}</math>. Since <math>\triangle BCD \sim \triangle ABC \sim \triangle CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=\frac{x^2}{\sqrt{x^2+1}}</math> and <math>BE=\frac{x}{\sqrt{x^2+1}}</math>. Let F be a point on <math>\overline{BC}</math> such that <math>\overline{EF}</math> is an altitude of triangle <math>CEB</math>. Note that <math>\triangle CEB \sim \triangle CFE \sim \triangle EFB</math>. Therefore, <math>BF=\frac{x}{x^2+1}</math> and <math>CF=\frac{x^3}{x^2+1}</math>. Since <math>\overline{CF}</math> and <math>\overline{BF}</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields: |
<cmath>[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))</cmath> | <cmath>[BEC]=[CED]=[BEA]=(x^3)/(2(x^2+1))</cmath> | ||
<cmath>[ABCD]=[AED]+[DEC]+[CEB]+[BEA]</cmath> | <cmath>[ABCD]=[AED]+[DEC]+[CEB]+[BEA]</cmath> | ||
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<cmath>(x)(x^2+1)=20x^3/(x^2+1)</cmath> | <cmath>(x)(x^2+1)=20x^3/(x^2+1)</cmath> | ||
<cmath>(x^2+1)^2=20x^2</cmath> | <cmath>(x^2+1)^2=20x^2</cmath> | ||
− | <cmath>x^4-18x^2+1=0 \implies x^2=9+ | + | <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> | ||
Revision as of 10:13, 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
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.