Difference between revisions of "1996 AHSME Problems/Problem 9"
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==Problem== | ==Problem== | ||
− | 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> | ||
− | |||
==Solution== | ==Solution== | ||
+ | === Solution 1 === | ||
+ | Since the two planes are perpendicular, it follows that <math>\triangle PAD</math> is a [[right triangle]]. Thus, <math>PD = \sqrt{PA^2 + AD^2} = \sqrt{PA^2 + AB^2} = \sqrt{34}</math>, which is option <math>\boxed{\text{B}}</math>. | ||
+ | === Solution 2 === | ||
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. | 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. | ||
Line 15: | Line 33: | ||
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>. | 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>. | ||
− | 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{B}</math> | + | 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{\text{B}}</math>. |
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=8|num-a=10}} | {{AHSME box|year=1996|num-b=8|num-a=10}} | ||
+ | |||
+ | [[Category:Introductory 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 |
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