Difference between revisions of "1979 AHSME Problems/Problem 23"
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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>. | 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>. | ||
− | + | <asy> | |
+ | size(150); | ||
+ | import patterns; | ||
+ | import olympiad; | ||
+ | pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux; | ||
+ | add("hatch",hatch()); | ||
+ | //AA=new A and etc. | ||
+ | draw(rotate(100,D)*(A--B--C--D--cycle)); | ||
+ | AA=rotate(100,D)*A; | ||
+ | BB=rotate(100,D)*D; | ||
+ | CC=rotate(100,D)*C; | ||
+ | DD=rotate(100,D)*B; | ||
+ | draw(BB--DD); | ||
+ | P=midpoint(AA--BB); | ||
+ | Q=midpoint(CC--DD); | ||
+ | draw(P--Q,dashed); | ||
+ | draw(P--CC,dashed); | ||
+ | draw(AA--CC,dashed); | ||
+ | dot(P); | ||
+ | dot(Q); | ||
+ | label("$A$",AA,W); | ||
+ | label("$B$",BB,S); | ||
+ | label("$C$",CC,E); | ||
+ | label("$D$",DD,N); | ||
+ | label("$P$",P,S); | ||
+ | label("$Q$",Q,E); | ||
+ | //Credit to TheMaskedMagician for the diagram | ||
+ | //Changes made by Treetor10145</asy> | ||
+ | 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>. | ||
+ | |||
+ | <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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | size(150); | ||
+ | import patterns; | ||
+ | import olympiad; | ||
+ | pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux,R; | ||
+ | add("hatch",hatch()); | ||
+ | //AA=new A and etc. | ||
+ | draw(rotate(100,D)*(A--B--C--D--cycle)); | ||
+ | AA=rotate(100,D)*A; | ||
+ | BB=rotate(100,D)*D; | ||
+ | CC=rotate(100,D)*C; | ||
+ | DD=rotate(100,D)*B; | ||
+ | draw(BB--DD); | ||
+ | P=midpoint(AA--BB); | ||
+ | Q=midpoint(CC--DD); | ||
+ | R=midpoint(AA--CC); | ||
+ | pair X=intersectionpoints(P--CC,BB--R)[0]; | ||
+ | draw(AA--CC,dashed); | ||
+ | draw(DD--X,dashed); | ||
+ | draw(X--CC,dashed); | ||
+ | draw(rightanglemark(CC,X,DD)); | ||
+ | dot(P); | ||
+ | dot(Q); | ||
+ | dot(X); | ||
+ | label("$A$",AA,W); | ||
+ | label("$B$",BB,S); | ||
+ | label("$C$",CC,E); | ||
+ | label("$D$",DD,N); | ||
+ | label("$P$",P,S); | ||
+ | label("$Q$",Q,E); | ||
+ | label("$D^\prime$",X,W); | ||
+ | //Credit to TheMaskedMagician for the diagram | ||
+ | //Changes made by Treetor10145</asy> | ||
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>. |
Revision as of 13:23, 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
.
is congruent to
, since
,
, and
are collinear.
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.