Difference between revisions of "1999 AHSME Problems/Problem 22"
(New page: == Problem == The graphs of <math>y = -|x-a| + b</math> and <math>y = |x-c| + d</math> intersect at points <math>(2,5)</math> and <math>(8,3)</math>. Find <math>a+c</math>. <math> \mathrm...) |
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Revision as of 14:35, 5 July 2013
Contents
Problem
The graphs of and intersect at points and . Find .
Solution
Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a 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:
Obviously, the maximum of the first graph is achieved when , and its value is . Similarly, the minimum of the other graph is . Therefore the two remaining vertices of the area between the graphs are and .
As the area has four right angles, it is a rectangle. Without actually computing and we can therefore conclude that .
Explanation of the last step
This is a property all rectangles in the coordinate plane have.
For a proof, note that for any rectangle its center can be computed as and at the same time as . In our case, we can compute that the center is , therefore , and .
An alternate last step
We can easily compute and using our picture.
Consider the first graph on the interval . The graph starts at height , then rises for steps to the height , and then falls for steps to the height . Solving for we get . Similarly we compute , therefore .
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 |
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