Difference between revisions of "1996 AHSME Problems/Problem 9"
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+ | ==Problem== | ||
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+ | Triangle <math>PAB</math> and square <math>ABCD</math> are in perpendicular planes. Given that <math>PA = 3, PB = 4</math> and <math>AB = 5</math>, what is <math>PD</math>? | ||
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+ | <math> \text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8 </math> | ||
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+ | ==Solution== | ||
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+ | Place the points on a coordinate grid, and let the <math>xy</math> plane (where <math>z=0</math>) contain triangle <math>PAB</math>. Square <math>ABCD</math> will have sides that are vertical. | ||
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+ | Place point <math>P</math> at <math>(0,0,0)</math>, and place <math>PA</math> on the x-axis so that <math>A(3,0,0)</math>, and thus <math>PA = 3</math>. | ||
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+ | Place <math>PB</math> on the y-axis so that <math>B(0,4,0)</math>, and thus <math>PB = 4</math>. This makes <math>AB = 5</math>, as it is the hypotenuse of a 3-4-5 right triangle (with the right angle being formed by the x and y axes). This is a clean use of the fact that <math>\triangle PAB</math> is a right triangle. | ||
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+ | Since <math>AB</math> is one side of <math>\square ABCD</math> with length <math>5</math>, <math>BC = 5</math> as well. Since <math>BC \perp AB</math>, and <math>BC</math> is also perpendicular to the <math>xy</math> plane, <math>BC</math> must run stright up and down. WLOG pick the up direction, and since <math>BC = 5</math>, we travel <math>5</math> units up to <math>C(0,4,5)</math>. Similarly, we travel <math>5</math> units up from <math>A(3,0,0)</math> to reach <math>D(3,0,5)</math>. | ||
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+ | We now have coordinates for <math>P</math> and <math>D</math>. The distance is <math>\sqrt{(5-0)^2 + (0-0)^2 + (3-0)^2} = \sqrt{34}</math>, which is option <math>\boxed{D}</math> | ||
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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=8|num-a=10}} | {{AHSME box|year=1996|num-b=8|num-a=10}} |
Revision as of 20:33, 18 August 2011
Problem
Triangle and square are in perpendicular planes. Given that and , what is ?
Solution
Place the points on a coordinate grid, and let the plane (where ) contain triangle . Square will have sides that are vertical.
Place point at , and place on the x-axis so that , and thus .
Place on the y-axis so that , and thus . This makes , as it is the hypotenuse of a 3-4-5 right triangle (with the right angle being formed by the x and y axes). This is a clean use of the fact that is a right triangle.
Since is one side of with length , as well. Since , and is also perpendicular to the plane, must run stright up and down. WLOG pick the up direction, and since , we travel units up to . Similarly, we travel units up from to reach .
We now have coordinates for and . The distance is , which is option
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |