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2002 AMC 8 Problems/Problem 20

Revision as of 16:15, 28 July 2013 by Totoro888 (talk | contribs) (→‎Solution 2)

Problem

The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$. Altitude $\overline{XC}$ bisects $\overline{YZ}$. What is the area (in square inches) of the shaded region?

[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,4), N); label(scale(0.8)*"$Y$", (0,0), W); label(scale(0.8)*"$Z$", (10,0), E); label(scale(0.8)*"$A$", (2.5,2.2), W); label(scale(0.8)*"$B$", (7.5,2.2), E); label(scale(0.8)*"$C$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy]

$\textbf{(A)}\ 1\frac{1}2\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 2\frac{1}2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 3\frac{1}2$

Solution 1

The shaded region is a right trapezoid. Assume WLOG that $YZ=8$. Then because the area of $\triangle XYZ$ is equal to 8, the height of the triangle $XC=2$. Because the line $AB$ is a midsegment, the top base of the triangle is $\frac12 AB = \frac14 YZ = 2$. Also, $AB$ divides $XC$ in two, so the height of the trapezoid is $\frac12 2 = 1$. The bottom base is $\frac12 YZ = 4$. The area of the shaded region is $\frac12 (2+4)(1) = \boxed{\text{(D)}\ 3}$.

Solution 2

Since $A$ and $B$ are the midpoints of $XY$ and $XZ$, respectively, $AY=AX=BX=BZ$. Draw segments $AC$ and $BC$. Since $XY=XZ$, it means that $X$ is on the perpendicular bisector of YZ. Then $YC=CZ$. $AB$ is the line that connects the midpoints of two sides of a triangle together, which means that $AB$ is parallel to and half in length of $YZ$. Then $AB=YC=CZ$. Since $AB$ is parallel to $YZ$, and $XY$ is the transversal, $\angle XAB=\angle AYC.$ Similarly, $\angle XBA=\angle BZC.$ Then, by SAS, $\triangle XAB=\triangle AYC=\triangle BZC$. Since corresponding parts of congruent triangles are congruent,$AC=BC=XA$. Using the fact that $AB$ is parallel to $YZ$, $\angle ABC=\angle BCZ$ and $\angle BAC=\angle ACY$. Also, $\angle ABC=\angle BCZ=\angle ACY$ because $\triangle ABC$ is isosceles. Now $\triangle XAB=\triangle AYC=\triangle BZC=\triangle ABC$. Draw an altitude through each of them such that each triangle is split into two congruent right triangles. Now there are a total of 8 congruent small triangles, each with area 1. The shaded area has three of these triangles, so it has area 3.

Basically the proof is to show $\triangle XAB=\triangle AYC=\triangle BZC=\triangle ABC$. If you just look at the diagram you can easily see that the triangles are congruent and you would solve this a lot faster. Anyways, the area of the shaded region is $\boxed{\text{(D)}\ 3}$.

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

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