Difference between revisions of "1979 AHSME Problems/Problem 23"
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\textbf{(E) }\frac{\sqrt{3}}{3}</math> | \textbf{(E) }\frac{\sqrt{3}}{3}</math> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | 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>. | ||
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Simplifying, <math>PQ^2=\frac{1}{2}</math>. | Simplifying, <math>PQ^2=\frac{1}{2}</math>. | ||
Therefore, <math>PQ=\frac{\sqrt{2}}{2}\Rightarrow</math> <math>\boxed{\textbf{C}}</math> | Therefore, <math>PQ=\frac{\sqrt{2}}{2}\Rightarrow</math> <math>\boxed{\textbf{C}}</math> | ||
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+ | Solution by treetor10145 | ||
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+ | ==Solution 2 (less overkill)== | ||
+ | Notice, like above said, that <math>P</math> is the midpoint of <math>AB</math> and <math>Q</math> is the midpoint of <math>CD</math>. | ||
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+ | To find the length of <math>PQ</math>, first draw in lines <math>CP</math> and <math>DP</math>. Notice that <math>DP</math> is an altitude of <math>\triangle ADP</math>. We find that <math>\angle{DAP} = 60 ^{\circ}</math> (since <math>\triangle ABD</math> is equilateral), and <math>AD=\frac{1}{2}</math>. Use the properties of 30-60-90 triangles to get <math>DP=\frac{\sqrt{3}}{2}</math>. Since <math>CP</math> is an altitude of a congruent equilateral triangle, <math>CP=DP=\frac{\sqrt{3}}{2}</math>. | ||
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+ | Notice that <math>\triangle CDP</math> is isosceles with <math>CP=DP</math>. Also, since <math>Q</math> is the midpoint of base <math>CD</math>, we can conclude that <math>PQ</math> is an altitude. We can use Pythagorean theorem to get the following (taking into consideration <math>DQ=\frac{1}{2}</math>): | ||
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+ | <cmath>DQ^2+PQ^2=PD^2</cmath> | ||
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+ | <cmath>\left(\frac{1}{2}\right)^2+PQ^2 = \left(\frac{\sqrt{3}}{2}\right)^2</cmath> | ||
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+ | <cmath>PQ^2=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}</cmath> | ||
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+ | <cmath>PQ=\frac{\sqrt{2}}{2}\Rightarrow \boxed{\textbf{C}}</cmath> | ||
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+ | -WannabeCharmander | ||
== See also == | == See also == |
Latest revision as of 21:35, 21 June 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 1
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
.
is congruent to
, since
,
, and
are collinear. Therefore, we can just find
.
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,
Solution by treetor10145
Solution 2 (less overkill)
Notice, like above said, that is the midpoint of
and
is the midpoint of
.
To find the length of , first draw in lines
and
. Notice that
is an altitude of
. We find that
(since
is equilateral), and
. Use the properties of 30-60-90 triangles to get
. Since
is an altitude of a congruent equilateral triangle,
.
Notice that is isosceles with
. Also, since
is the midpoint of base
, we can conclude that
is an altitude. We can use Pythagorean theorem to get the following (taking into consideration
):
-WannabeCharmander
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 |
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