Difference between revisions of "2021 AMC 12B Problems/Problem 17"
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Revision as of 15:01, 11 February 2021
Problem 17
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
![\[14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 4\cdot\frac rs + 2\cdot\frac sr.\]](http://latex.artofproblemsolving.com/4/d/1/4d1b56b25fb1019e002964ad8b94413da85b14f3.png)
Thus, the ratio 
satisfies 
; solving yields 
.
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by 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. 
