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

1999 AHSME Problems/Problem 22

Problem

The graphs of $y = -|x-a| + b$ and $y = |x-c| + d$ intersect at points $(2,5)$ and $(8,3)$. Find $a+c$.

$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18$

Solution

Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a $45^\circ$ angle, in the second graph they point upwards. One can easily find out that the only way how to get these graphs to intersect in two points is the one depicted below:

[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label("$(2,5)$",X,W*1.5); label("$(8,3)$",Y,E*1.5); [/asy]

Obviously, the maximum of the first graph is achieved when $x=a$, and its value is $-0+b=b$. Similarly, the minimum of the other graph is $(c,d)$. Therefore the two remaining vertices of the area between the graphs are $(a,b)$ and $(c,d)$.

As the area has four right angles, it is a rectangle. Without actually computing $a$ and $c$ we can therefore conclude that $a+c=2+8=\boxed{10}$.

Explanation of the last step

This is a property all rectangles in the coordinate plane have.

For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\left(\frac{2+8}2,\frac{5+3}2\right)=(5,4)$, therefore $\frac{a+c}2=5$, and $a+c=10$.

[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label("$(2,5)$",X,W*1.5); label("$(8,3)$",Y,E*1.5); label("$(a,b)$",(4,7),N); label("$(c,d)$",(6,1),S); [/asy]

An alternate last step

We can easily compute $a$ and $c$ using our picture.

[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label("$(2,5)$",X,W*1.5); label("$(8,3)$",Y,E*1.5); label("$(a,b)$",(4,7),N); label("$(c,d)$",(6,1),S); [/asy]

Consider the first graph on the interval $[2,8]$. The graph starts at height $5$, then rises for $a-2$ steps to the height $b=5+(a-2)$, and then falls for $8-a$ steps to the height $3=5+(a-2)-(8-a)$. Solving for $a$ we get $a=4$. Similarly we compute $c=6$, therefore $a+c=10$.

See also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1999_AHSME_Problems/Problem_22&oldid=56583"