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

Revision as of 22:02, 28 December 2023

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 $\overline{YZ}$ . The area (in square inches) of the shaded region is

$[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$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 trapezoid 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 $BC$ . Drawing an altitude in an isoceles triangle splits the triangle into 2 congruent triangles and we also know that $YC=CZ$ . $AB$ 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, $AC=BC=XA$ . Since ACY and BCZ are now isosceles triangles, $\angle AYC=\angle ACY$ and $\angle BZC=\angle BCZ$ Using the fact that $AB$ is parallel to $YZ$ , $\angle ACY=\angle CAB$ and $\angle BCZ=\angle CBA$ . 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 $\boxed{\text{(D)}\ 3}$ .

Solution 3

We know the area of triangle $XYZ$ is $8$ square inches. The area of a triangle can also be represented as $\frac{bh}{2}$ or in this problem $\frac{XC\cdot YZ}{2}$ . By solving, we have $\frac{XC\cdot YZ}{2} = 8,$ $XC\cdot YZ = 16.$

With SAS congruence, triangles $XCY$ and $XCZ$ are congruent. Hence, triangle $XCY = \frac{8}{2} = 4$ . (Let's say point $D$ is the intersection between line segments $XC$ and $AB$ .) We can find the area of the trapezoid $ADCY$ by subtracting the area of triangle $XAD$ from $4$ .

We find the area of triangle $XAD$ by the $\frac{bh}{2}$ formula- $\frac{XD\cdot AD}{2} = \frac{\frac{XC}{2}\cdot AD}{2}$ . $AD$ is $\frac{1}{4}$ of $YZ$ from solution 1. The area of $XAD$ is $\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1$ .

Therefore, the area of the shaded area- trapezoid $ADCY$ has area $4-1 = \boxed{\text{(D)}\ 3}$ .

- sarah07

Solution 4 (Dummed down)

The area of triangle $XYZ$ is $8$ square inches. You can turn the triangle into a rectangle by drawing a triangle on the left side with the hypotenuse of it being $YAX$ and another triangle on the right side, with its hypotenuse being $XBZ$ . After drawing the square, you can cut it into $8$ squares. The area of the rectangle is $16$ square inches because the triangle on the left is half of $XYCZ$ and there's another triangle on the other side, equaling them to $8$ square inches. We will be focusing on the left side of the rectangle of $XC$ because it includes the shaded region. It's split into $4$ equal squares. The area of this triangle is $8$ square inches because the total area of the rectangle is $16$ square inches and $16$ divided by $2$ is $8$ . There are $4$ sections, so you would do $8$ divided by $4$ in order to find the area of one square. That means that the area of the top right square is $2$ inches and because it's not needed, we will subtract $2$ from $8$ to get rid of it. If you turn around the paper, you notice that the square with half of the shaded region, and the square above it is the shaded region, except flipped and turned. Therefore, the remaining of the area of it which is $6$ should then be divided by $2$ in order to find the shaded region since the shaded region is equal to the other square and half a square. $6$ divided by $2$ is $3$ , so the answer is $3$ .

Solution 5 (Dividing into Congruent Triangles)

Super fast after convincing yourself that they are congruent

$[asy] /* AMC8 2002 #20 Problem */ draw((0,0)--(10,0)--(5,4)--cycle); draw((2.5,2)--(7.5,2)); draw((5,4)--(5,0)); fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); label(scale(0.8)*"$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$", (5,0), S); fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white); /* Edit for solution */ /* A--C */ draw((2.5,2)--(5,0)); /* A--[half of YC] */ draw((2.5,2)--(2.5,0)); label(scale(0.8)*"$D$",(2.5,0),S); /* B--C */ draw((7.5,2)--(5,0)); /* B--[half of CZ] */ draw((7.5,2)--(7.5,0)); label(scale(0.8)*"$E$",(7.5,0),S); /* drawing right angles */ draw((2.5,0.5)--(2,0.5)); draw((2,0.5)--(2,0)); draw((2.5,0.5)--(3,0.5)); draw((3,0.5)--(3,0)); draw((7.5,0.5)--(7,0.5)); draw((7,0.5)--(7,0)); draw((7.5,0.5)--(8,0.5)); draw((8,0.5)--(8,0)); draw((5.5,2.5)--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--cycle); label(scale(0.8)*"$F$", (5,2), NE); label(scale(1)*"All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section.", (5,-1), S); [/asy]$

We draw the triangles as shown above. What seems like a half-square (or a corner of a square) indicates $90$ °. (This is not a random idea - the two original two triangles on the top inspires us to do so.)

Note that $\overline{AD}$ is perpendicular to $\overline{YC}$ ; $\overline{BE}$ is perpendicular to $\overline{CZ}$ .

$\overline{AB}$ is perpendicular to $\overline{YZ}$ , so because of corresponding angles theorem, $\angle YAD=\angle AXF$ . We are also told $\overline{YA}=\overline{AX}$ . Both $\bigtriangleup \text{XAF}$ and $\bigtriangleup \text{AYD}$ have a $90$ ° angle. Because of AAS Congruency (Angle-Angle-Side; $90$ °, $\angle YAD$ , and side $\overline{YA}$ ), $\bigtriangleup \text{XAF}$ is congruent to $\bigtriangleup \text{AYD}$ . Because of symmetry, the same goes for $\bigtriangleup \text{XFB}$ and $\bigtriangleup \text{BEZ}$ .

$\bigtriangleup \text{YAD}$ is congruent to $\bigtriangleup \text{CAD}$ , which is congruent to $\bigtriangleup \text{ACF}$ because of symmetry. So there are $4$ congruent triangles on the left side. There is also $4$ congruent triangles on the right side because of symmetry.

There are $8$ congruent triangles in all. We are told the area of the triangle $\bigtriangleup \text{XYZ}$ is 8. $8\div8=1$ , so each of the small congruent triangles is $1$ . The shaded area contains $3$ small triangles, so the area of the shaded section is $1\cdot3=\boxed{\text{(D) }3}$ .

- JoyfulSapling

Video Solution

https://www.youtube.com/watch?v=zwy5U5IQi88 ~David

Revision as of 22:01, 28 December 2023 (view source) Joyfulsapling (talk \| contribs) (→‎Super fast after convincing yourself that they are congruent) ← Older edit		Revision as of 22:02, 28 December 2023 (view source) Joyfulsapling (talk \| contribs) m (→‎Super fast after convincing yourself that they are congruent) Newer edit →
Line 97:		Line 97:
	</asy>		</asy>

−	We draw the triangles as shown above. What seems like a half-square (or a corner of a square) indicates <math>9</math>~~<math>0</math>•~~(This is not a random idea - the two original two triangles on the top inspires us to do so.)	+	We draw the triangles as shown above. What seems like a half-square (or a corner of a square) indicates <math>90</math>°. (This is not a random idea - the two original two triangles on the top inspires us to do so.)

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

Page

Toolbox

Search