Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
 
(41 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
The area of triangle <math>XYZ</math> is 8 square inches. Points <math>A</math> and <math>B</math> are midpoints of congruent segments <math> \overline{XY} </math> and <math> \overline{XZ} </math>. Altitude <math> \overline{XC} </math> bisects <math> \overline{YZ} </math>. What is the area (in square inches) of the shaded region?
+
The area of triangle <math>XYZ</math> is 8 square inches. Points <math>A</math> and <math>B</math> are midpoints of congruent segments <math>\overline{XY}</math> and <math>\overline{XZ}</math>. Altitude <math>\overline{XC}</math> bisects <math>\overline{YZ}</math>. The area (in square inches) of the shaded region is
  
 
<asy>
 
<asy>
Line 19: Line 19:
  
 
==Solution 1==
 
==Solution 1==
−
The shaded region is a right trapezoid. Assume WLOG that <math>YZ=8</math>. Then because the area of <math>\triangle XYZ</math> is equal to 8, the height of the triangle <math>XC=2</math>. Because the line <math>AB</math> is a midsegment, the top base of the triangle is <math>\frac12 AB = \frac14 YZ = 2</math>. Also, <math>AB</math> divides <math>XC</math> in two, so the height of the trapezoid is <math>\frac12 (2) = 1</math>. The bottom base is <math>\frac12 YZ = 4</math>. The area of the shaded region is <math>\frac12 (2+4)(1) = \boxed{\text{(D)}\ 3}</math>.
+
We know the area of triangle <math>XYZ</math> is <math>8</math> square inches. The area of a triangle can also be represented as <math>\frac{bh}{2}</math> or in this problem <math>\frac{XC\cdot YZ}{2}</math>. By solving, we have <cmath>\frac{XC\cdot YZ}{2} = 8,</cmath> <cmath>XC\cdot YZ = 16.</cmath>  
  
−
==Solution 2==
+
With SAS congruence, triangles <math>XCY</math> and <math>XCZ</math> are congruent. Hence, triangle <math>XCY = \frac{8}{2} = 4</math>. (Let's say point <math>D</math> is the intersection between line segments <math>XC</math> and <math>AB</math>.) We can find the area of the trapezoid <math>ADCY</math> by subtracting the area of triangle <math>XAD</math> from <math>4</math>.  
−
Since <math>A</math> and <math>B</math> are the midpoints of <math>XY</math> and <math>XZ</math>, respectively, <math>AY=AX=BX=BZ</math>.
 
−
Draw segments <math>AC</math> and <math>BC</math>.  
 
−
Since <math>XY=XZ</math>, it means that <math>X</math> is on the perpendicular bisector of YZ. Then <math>YC=CZ</math>.
 
−
<math>AB</math> is the line that connects the midpoints of two sides of a triangle together, which means that <math>AB</math> is parallel to and half in length of <math>YZ</math>. Then <math>AB=YC=CZ</math>.
 
−
Since <math>AB</math> is parallel to <math>YZ</math>, and <math>XY</math> is the transversal, <math>\angle XAB=\angle AYC.</math> Similarly, <math>\angle XBA=\angle BZC.</math> Then, by SAS, <math>\triangle XAB=\triangle AYC=\triangle BZC</math>.
 
−
Since corresponding parts of congruent triangles are congruent,<math>AC=BC=XA</math>.
 
−
Using the fact that <math>AB</math> is parallel to <math>YZ</math>, <math>\angle ABC=\angle BCZ</math> and <math>\angle BAC=\angle ACY</math>. Also, <math>\angle ABC=\angle BCZ=\angle ACY</math> because <math>\triangle ABC</math> is isosceles.
 
−
Now <math>\triangle XAB=\triangle AYC=\triangle BZC=\triangle ABC</math>.
 
−
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.
 
  
 +
We find the area of triangle <math>XAD</math> by the <math>\frac{bh}{2}</math> formula- <math>\frac{XD\cdot AD}{2} = \frac{\frac{XC}{2}\cdot AD}{2}</math>. <math>AD</math> is <math>\frac{1}{4}</math> of <math>YZ</math> from solution 1. The area of <math>XAD</math> is <cmath>\frac{\frac{XC}{2}\cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1</cmath>.
  
−
Basically the proof is to show <math>\triangle XAB=\triangle AYC=\triangle BZC=\triangle ABC</math>. 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 <math>\boxed{\text{(D)}\ 3}</math>.
+
Therefore, the area of the shaded area- trapezoid <math>ADCY</math> has area <math>4-1 = \boxed{\text{(D)}\ 3}</math>.  
 +
 
 +
 
 +
- sarah07
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=zwy5U5IQi88  ~David
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/wIVHVxhA7xg
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=19|num-a=21}}
 
{{AMC8 box|year=2002|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:33, 29 October 2024

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

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

Video Solution

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

Video Solution by WhyMath

https://youtu.be/wIVHVxhA7xg

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_8_Problems/Problem_20&oldid=230847"