Difference between revisions of "1979 AHSME Problems/Problem 23"
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Treetor10145 (talk | contribs) (Small changes to make solution more clear) |
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− | Next, we need to find <math>\cos(\angle | + | Next, we need to find <math>\cos(\angle PCQ)</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>. |
− | <math>\angle PCD</math> is congruent to <math>\angle D^\prime CD</math>, since <math>P</math>, <math>D^\prime</math>, and <math>C</math> are collinear. | + | <math>\angle PCD</math> is congruent to <math>\angle D^\prime CD</math>, since <math>P</math>, <math>D^\prime</math>, and <math>C</math> are collinear. Therefore, we can just find <math>\cos(\angle D^\prime CD)</math>. |
Note that <math>\triangle CD^\prime D</math> is a right triangle with <math>\angle CD^\prime D</math> as a right angle. | Note that <math>\triangle CD^\prime D</math> is a right triangle with <math>\angle CD^\prime D</math> as a right angle. | ||
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//Changes made by Treetor10145</asy> | //Changes made by Treetor10145</asy> | ||
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+ | 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>. | 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 | + | Since <math>\cos(\angle D^\prime CD)=\cos(\angle PCQ)=\frac{CD^\prime}{CD}</math>, <math>\cos(\angle PCQ)=\frac{\frac{\sqrt{3}}{3}}{1}=\frac{\sqrt{3}}{3}</math> |
− | <cmath>PQ^2=CP^2+CQ^2-2(CP)(CQ)\cos(\angle | + | <cmath>PQ^2=CP^2+CQ^2-2(CP)(CQ)\cos(\angle PCQ)</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> | <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>. | Simplifying, <math>PQ^2=\frac{1}{2}</math>. | ||
− | Therefore, <math>PQ=\frac{\sqrt{2}}{2}</math> | + | Therefore, <math>PQ=\frac{\sqrt{2}}{2}\Rightarrow</math> <math>\boxed{\textbf{C}}</math> |
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− | <math>\boxed{\textbf{C}}</math> | ||
== See also == | == See also == |
Revision as of 14:29, 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
.
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,
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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