Difference between revisions of "1999 AHSME Problems/Problem 22"

Revision as of 16:18, 19 July 2020

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 1

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 $a+c=10$ .

Solution 2

Plug the coordinates into the two equations and set the equations equal to each other. The $b$ and $d$ terms cancel out and you're left with just the following.

$|3-a| - |2-a| = |2-c| - |3-c|$

If we solve individually for the values of $a$ by setting both absolute values on the right to positive and setting one absolute value on the left to a negative, $a = 4$ . Doing the same with $c$ , we see that $c = 6$ . So, $a+c=10$ .

@@ Line 62: / Line 62: @@
 <math>|3-a| - |2-a| = |2-c| - |3-c|</math>
-If we solve individually for the values of <math>a</math> by setting both absolute values on the right to positive and setting one absolute value on the left to a negative, <math>a = 5</math>. Doing the same with <math>c</math>, we see that <math>c = 5</math> as well. So, <math>a+c=10</math>.
+If we solve individually for the values of <math>a</math> by setting both absolute values on the right to positive and setting one absolute value on the left to a negative, <math>a = 4</math>. Doing the same with <math>c</math>, we see that <math>c = 6</math>. So, <math>a+c=10</math>.
 == See also ==
 {{AHSME box|year=1999|num-b=21|num-a=23}}
 {{MAA Notice}}

1999 AHSME (Problems • Answer Key • Resources)
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