Difference between revisions of "1979 AHSME Problems/Problem 23"
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==Solution== | ==Solution== | ||
+ | Note that the distance <math>PQ</math> will be minimized when <math>P</math> is the midpoint of <math>AB</math> and <math>Q</math> is the midpoint of <math>CD</math>. | ||
+ | |||
+ | To find this distance, consider triangle <math>\triangle PCQ</math>. <math>Q</math> is the midpoint of <math>CD</math>, so <math>CQ=\frac{1}{2}</math>. Additionally, since <math>CP</math> is the altitude of equilateral <math>\triangle ABC</math>, <math>CP=\frac{\sqrt{3}}{2}</math>. | ||
+ | |||
+ | Next, we need to find <math>\cos(\angle PCD)</math> in order to find <math>PQ</math> by the Law of Cosines. To do so, drop down <math>D</math> onto <math>\triangle ABC</math> to get the point <math>D^\prime</math>. Note that <math>\triangle CD^\prime D</math> is a right triangle with <math>\angle CD^\prime D</math> as a right angle. As given by the problem, <math>CD=1</math>. | ||
+ | |||
+ | |||
+ | Note that <math>D^\prime</math> is the centroid of equilateral <math>\triangle ABC</math>. Additionally, since <math>\triangle ABC</math> is equilateral, <math>D^\prime</math> is also the orthocenter. Due to this, the distance from <math>C</math> to <math>D^\prime</math> is <math>\frac{2}{3}</math> of the altitude of <math>\triangle ABC</math>. Therefore, <math>CD^\prime=\frac{\sqrt{3}}{3}</math>. | ||
+ | |||
+ | Since <math>\cos(\angle PCD)=\frac{CD^\prime}{CD}</math>, <math>\cos(\angle PCD)=\frac{\frac{\sqrt{3}}{3}}{1}=\frac{\sqrt{3}}{3}</math> | ||
+ | |||
+ | <cmath>PQ^2=CP^2+CQ^2-2(CP)(CQ)\cos(\angle PCD)</cmath> | ||
+ | <cmath>PQ^2=\frac{3}{4}+\frac{1}{4}-2\left(\frac{\sqrt{3}}{4}\right)\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{3}\right)</cmath> | ||
+ | Simplifying, <math>PQ^2=\frac{1}{2}</math>. | ||
+ | Therefore, <math>PQ=\frac{\sqrt{2}}{2}</math> | ||
+ | |||
<math>\boxed{\textbf{C}}</math> | <math>\boxed{\textbf{C}}</math> | ||
Revision as of 09:23, 12 February 2018
Problem 23
The edges of a regular tetrahedron with vertices , and
each have length one.
Find the least possible distance between a pair of points
and
, where
is on edge
and
is on edge
.
Solution
Note that the distance will be minimized when
is the midpoint of
and
is the midpoint of
.
To find this distance, consider triangle .
is the midpoint of
, so
. Additionally, since
is the altitude of equilateral
,
.
Next, we need to find in order to find
by the Law of Cosines. To do so, drop down
onto
to get the point
. Note that
is a right triangle with
as a right angle. As given by the problem,
.
Note that is the centroid of equilateral
. Additionally, since
is equilateral,
is also the orthocenter. Due to this, the distance from
to
is
of the altitude of
. Therefore,
.
Since ,
Simplifying,
.
Therefore,
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.