Difference between revisions of "2021 AMC 12B Problems/Problem 17"
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Let <math>ABCD</math> be an isoceles trapezoid having parallel bases <math>\overline{AB}</math> and <math>\overline{CD}</math> with <math>AB>CD.</math> Line segments from a point inside <math>ABCD</math> to the vertices divide the trapezoid into four triangles whose areas are <math>2, 3, 4,</math> and <math>5</math> starting with the triangle with base <math>\overline{CD}</math> and moving clockwise as shown in the diagram below. What is the ratio <math>\frac{AB}{CD}?</math> | Let <math>ABCD</math> be an isoceles trapezoid having parallel bases <math>\overline{AB}</math> and <math>\overline{CD}</math> with <math>AB>CD.</math> Line segments from a point inside <math>ABCD</math> to the vertices divide the trapezoid into four triangles whose areas are <math>2, 3, 4,</math> and <math>5</math> starting with the triangle with base <math>\overline{CD}</math> and moving clockwise as shown in the diagram below. What is the ratio <math>\frac{AB}{CD}?</math> | ||
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<asy> | <asy> | ||
unitsize(100); | unitsize(100); | ||
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dot(A^^B^^C^^D^^P); | dot(A^^B^^C^^D^^P); | ||
</asy> | </asy> | ||
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<math>\textbf{(A)}\: 3\qquad\textbf{(B)}\: 2+\sqrt{2}\qquad\textbf{(C)}\: 1+\sqrt{6}\qquad\textbf{(D)}\: 2\sqrt{3}\qquad\textbf{(E)}\: 3\sqrt{2}</math> | <math>\textbf{(A)}\: 3\qquad\textbf{(B)}\: 2+\sqrt{2}\qquad\textbf{(C)}\: 1+\sqrt{6}\qquad\textbf{(D)}\: 2\sqrt{3}\qquad\textbf{(E)}\: 3\sqrt{2}</math> | ||
Revision as of 18:09, 15 April 2021
Contents
Problem
Let 
be an isoceles trapezoid having parallel bases 
and 
with 
Line segments from a point inside to the vertices divide the trapezoid into four triangles whose areas are 
and 
starting with the triangle with base and moving clockwise as shown in the diagram below. What is the ratio 
![[asy] unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.3, 0.9), D=(-0.3, 0.9), P=(0.2, 0.5), E=(0.1, 0.75), F=(0.4, 0.5), G=(0.15, 0.2), H=(-0.3, 0.5); draw(A--B--C--D--cycle, black); draw(A--P, black); draw(B--P, black); draw(C--P, black); draw(D--P, black); label("$A$",A,(-1,0)); label("$B$",B,(1,0)); label("$C$",C,(1,-0)); label("$D$",D,(-1,0)); label("$2$",E,(0,0)); label("$3$",F,(0,0)); label("$4$",G,(0,0)); label("$5$",H,(0,0)); dot(A^^B^^C^^D^^P); [/asy]](http://latex.artofproblemsolving.com/2/9/c/29c42141e2c79d3a16294b55a9be52158c6f9681.png)
Solution
Without loss let 
have vertices 
, 
, 
, and 
, with 
and 
. Also denote by 
the point in the interior of .
Let 
and 
be the feet of the perpendiculars from to 
and 
, respectively. Observe that 
and 
. Now using the formula for the area of a trapezoid yields
![\[14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 2\cdot\frac rs + 4\cdot\frac sr.\]](http://latex.artofproblemsolving.com/9/e/0/9e00a7563e000cf1b9723fcc209e423a9ad51a4e.png)
Thus, the ratio 
satisfies 
; solving yields 
.
Solution 2
Let 
be the bottom base, 
be the top base, 
be the height of the bottom triangle, 
be the height of the top triangle.
Thus, 
so 
Let 
so we get 
This gives us a quadratic in 
ie. 
so 
- Solution by MathAwesome123, added by ccx09
Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area)
~ pi_is_3.14
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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. 
