Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, | + | Quadrilateral <math>ABCD</math> has right angles at <math>B</math> and <math>C</math>, <math>\triangle ABC \sim \triangle BCD</math>, and <math>AB > BC</math>. There is a point <math>E</math> in the interior of <math>ABCD</math> such that <math>\triangle ABC \sim \triangle CEB</math> and the area of <math>\triangle AED</math> is <math>17</math> times the area of <math>\triangle CEB</math>. What is <math>\tfrac{AB}{BC}</math>? |
+ | |||
<math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math> | <math>\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } 2 + \sqrt{2} \qquad \textbf{(C) } \sqrt{17} \qquad \textbf{(D) } 2 + \sqrt{5} \qquad \textbf{(E) } 1 + 2\sqrt{3}</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | 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>[ABCD]=[AED]+[DEC]+[CEB]+[BEA]</cmath> | ||
+ | <cmath>(AB+CD)(BC)/2= 17*[CEB]+ [CEB] + [CEB] + [CEB]</cmath> | ||
+ | <cmath>(x^3+x)/2=(20x^3)/(2(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^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> | ||
− | + | ||
− | <math> | + | ==Solution 2== |
− | + | Draw line <math>FG</math> through <math>E</math>, with <math>F</math> on <math>BC</math> and <math>G</math> on <math>AD</math>, <math>FG \parallel AB</math>. WLOG let <math>CD=1</math>, <math>CB=x</math>, <math>AB=x^2</math>. By weighted average <math>FG=\frac{1+x^4}{1+x^2}</math>. | |
− | + | ||
− | < | + | Meanwhile, <math>FE:EG=[\triangle CBE]:[\triangle ADE]=1:17</math>. This follows from comparing the ratios of triangle DEG to CFE and triangle AEG to FEB, both pairs in which the two triangles share a height perpendicular to FG, and have base ratio <math>EG:FE</math>. |
− | + | ||
− | <math>( | + | <math>FE=\frac{x^2}{1+x^2}</math>. We obtain <math>\frac{1+x^4}{1+x^2}=\frac{18x^2}{1+x^2}</math>, |
− | <math> | + | namely <math>x^4-18x^2+1=0</math>. |
− | <math> | + | |
− | <math> | + | The rest is the same as Solution 1. |
− | + | ||
+ | == Solution 3== | ||
+ | Let <math>A=(-1,4a), B=(-1,0), C=(1,0), D=\bigg(1,\frac{1}{a}\bigg)</math>. Then from the similar triangles condition, we compute <math>CE=\frac{4a}{\sqrt{4a^2+1}}</math> and <math>BE=\frac{2}{\sqrt{4a^2+1}}</math>. Hence, the <math>y</math>-coordinate of <math>E</math> is just <math>\frac{BE\cdot CE}{BC}=\frac{4a}{4a^2+1}</math>. Since <math>E</math> lies on the unit circle, we can compute the <math>x</math> coordinate as <math>\frac{1-4a^2}{4a^2+1}</math>. By Shoelace, we want <cmath>\frac{1}{2}\det\begin{bmatrix} | ||
+ | -1 & 4a & 1\\ | ||
+ | \frac{1-4a^2}{4a^2+1} & \frac{4a}{4a^2+1} & 1\\ | ||
+ | 1 & \frac{1}{a} & 1 | ||
+ | \end{bmatrix}=17\cdot\frac{1}{2}\cdot 2 \cdot \frac{4a}{4a^2+1}</cmath>Factoring out denominators and expanding by minors, this is equivalent to | ||
+ | <cmath>\frac{32a^4-8a^2+2}{2a(4a^2+1)}=\frac{68a}{4a^2+1} \Longrightarrow 16a^4-72a^2+1=0</cmath>This factors as <math>(4a^2-8a-1)(4a^2+8a-1)=0</math>, so <math>a=1+\frac{\sqrt{5}}{2}</math> and so the answer is <math> \textbf{(D) \ }</math>. | ||
+ | |||
+ | == Notes== | ||
+ | 1) <math>\sqrt{17}</math> is the most relevant answer choice because it shares numbers with the givens of the problem. | ||
+ | |||
+ | 2) It's a very good guess to replace finding the area of triangle AED with the area of the triangle DAF, where F is the projection of D onto AB(then find the closest answer choice). | ||
==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}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 13:56, 15 February 2021
Problem
Quadrilateral has right angles at and , , and . There is a point in the interior of such that and the area of is times the area of . What is ?
Solution 1
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 through , with on and on , . WLOG let , , . By weighted average .
Meanwhile, . This follows from comparing the ratios of triangle DEG to CFE and triangle AEG to FEB, both pairs in which the two triangles share a height perpendicular to FG, and have base ratio .
. We obtain , namely .
The rest is the same as Solution 1.
Solution 3
Let . Then from the similar triangles condition, we compute and . Hence, the -coordinate of is just . Since lies on the unit circle, we can compute the coordinate as . By Shoelace, we want Factoring out denominators and expanding by minors, this is equivalent to This factors as , so and so the answer is .
Notes
1) is the most relevant answer choice because it shares numbers with the givens of the problem.
2) It's a very good guess to replace finding the area of triangle AED with the area of the triangle DAF, where F is the projection of D onto AB(then find the closest answer choice).
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 |
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