2010 AMC 8 Problem 17

Attempt Problem 17 of the 2010 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

The diagram shows an octagon consisting of 1010 unit squares. The portion below PQ‾\overline{PQ} is a unit square and a triangle with base 5.5. If PQ‾\overline{PQ} bisects the area of the octagon, what is the ratio XQQY?\dfrac{XQ}{QY}?
25 \dfrac{2}{5}
12 \dfrac{1}{2}
35 \dfrac{3}{5}
23 \dfrac{2}{3}
34 \dfrac{3}{4}
Answer: D
Concepts:area decompositiontriangle area
Small hint:
Below PQPQ, subtract the unit square to get a triangle area equation.
Big hint:
The line PQPQ cuts the octagon into two equal areas.
Solution:
Since PQPQ bisects the area, the area under the line is 5.5. Removing the square on the right makes the bottom a triangle of base 55 with area 4.4. Let the base of this triangle be PZ.PZ.
The area being 44 means (PZ)(ZQ)2=5(ZQ)2=4\dfrac{(PZ)(ZQ)}{2}=\frac{5(ZQ)}{2} = 4   ⟹  ZQ=1.6.\implies ZQ = 1.6.
Therefore, QY=QZ−1=0.6QY = QZ-1 = 0.6 and XQ=2−QZ=0.4. XQ = 2-QZ = 0.4. This would make XQQY=0.40.6=23.\dfrac{XQ}{QY} = \dfrac {0.4}{0.6} = \dfrac 23.
Thus, the answer is D .
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