Difference between revisions of "1999 AHSME Problems/Problem 22"
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<math> \mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18</math> | <math> \mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18</math> | ||
− | == Solution == | + | == Solution 1== |
Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a <math>45^\circ</math> 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: | Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a <math>45^\circ</math> 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: | ||
Line 56: | Line 56: | ||
Consider the first graph on the interval <math>[2,8]</math>. The graph starts at height <math>5</math>, then rises for <math>a-2</math> steps to the height <math>b=5+(a-2)</math>, and then falls for <math>8-a</math> steps to the height <math>3=5+(a-2)-(8-a)</math>. Solving for <math>a</math> we get <math>a=4</math>. Similarly we compute <math>c=6</math>, therefore <math>a+c=10</math>. | Consider the first graph on the interval <math>[2,8]</math>. The graph starts at height <math>5</math>, then rises for <math>a-2</math> steps to the height <math>b=5+(a-2)</math>, and then falls for <math>8-a</math> steps to the height <math>3=5+(a-2)-(8-a)</math>. Solving for <math>a</math> we get <math>a=4</math>. Similarly we compute <math>c=6</math>, therefore <math>a+c=10</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Plug the coordinates into the two equations and set the equations equal to each other. The <math>b</math> and <math>d</math> terms cancel out and you're left with just the following. | ||
+ | |||
+ | <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 = 4</math>. Doing the same with <math>c</math>, we see that <math>c = 6</math>. So, <math>a+c=10</math>. | ||
== See also == | == See also == | ||
{{AHSME box|year=1999|num-b=21|num-a=23}} | {{AHSME box|year=1999|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:02, 17 October 2020
Contents
Problem
The graphs of and intersect at points and . Find .
Solution 1
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 .
Solution 2
Plug the coordinates into the two equations and set the equations equal to each other. The and terms cancel out and you're left with just the following.
If we solve individually for the values of by setting both absolute values on the right to positive and setting one absolute value on the left to a negative, . Doing the same with , we see that . So, .
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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