Difference between revisions of "2002 AMC 8 Problems/Problem 20"

Revision as of 13:30, 26 October 2016

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?

$[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(4,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$X$", (5,6), N); label(scale(0.8)*"$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, since those triangles are congruent, you can split each in half to find eight congruent triangles with area 1, and since the shaded region has three of these triangles, its area is $\boxed{\text{(D)}\ 3}$ .

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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Difference between revisions of "2002 AMC 8 Problems/Problem 20"

Revision as of 13:30, 26 October 2016

Contents

Problem

Solution 1

Solution 2

See Also

@@ Line 7: / Line 7: @@
 draw((2.5,2)--(7.5,2));
 draw((5,4)--(5,0));
-fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey);
+fill((0,0)--(2.5,2)--(4,2)--(5,0)--cycle, mediumgrey);
-label(scale(0.8)*"$X$", (5,4), N);
+label(scale(0.8)*"$X$", (5,6), N);
 label(scale(0.8)*"$Y$", (0,0), W);
 label(scale(0.8)*"$Z$", (10,0), E);