Difference between revisions of "1996 AHSME Problems/Problem 9"
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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>? | 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>? | ||
+ | <asy> | ||
+ | real r=sqrt(2)/2; | ||
+ | draw(origin--(8,0)--(8,-1)--(0,-1)--cycle); | ||
+ | draw(origin--(8,0)--(8+r, r)--(r,r)--cycle); | ||
+ | filldraw(origin--(-6*r, -6*r)--(8-6*r, -6*r)--(8, 0)--cycle, white, black); | ||
+ | filldraw(origin--(8,0)--(8,6)--(0,6)--cycle, white, black); | ||
+ | pair A=(6,0), B=(2,0), C=(2,4), D=(6,4), P=B+1*dir(-65); | ||
+ | draw(A--P--B--C--D--cycle); | ||
+ | dot(A^^B^^C^^D^^P); | ||
+ | label("$A$", A, dir((4,2)--A)); | ||
+ | label("$B$", B, dir((4,2)--B)); | ||
+ | label("$C$", C, dir((4,2)--C)); | ||
+ | label("$D$", D, dir((4,2)--D)); | ||
+ | label("$P$", P, dir((4,2)--P)); | ||
+ | </asy> | ||
+ | <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== | ==Solution== | ||
=== Solution 1 === | === Solution 1 === | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
[[Category:3D Geometry Problems]] | [[Category:3D Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:07, 5 July 2013
Problem
Triangle and square
are in perpendicular planes. Given that
and
, what is
?
Solution
Solution 1
Since the two planes are perpendicular, it follows that is a right triangle. Thus,
, which is option
.
Solution 2
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.