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== |
− | <math>\boxed{\textbf{ | + | 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>. | ||
+ | |||
+ | <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 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. 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. | ||
+ | |||
+ | |||
+ | <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> | ||
+ | |||
+ | |||
+ | 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 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 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> | ||
+ | Simplifying, <math>PQ^2=\frac{1}{2}</math>. | ||
+ | Therefore, <math>PQ=\frac{\sqrt{2}}{2}\Rightarrow</math> <math>\boxed{\textbf{C}}</math> | ||
+ | |||
+ | Solution by treetor10145 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | |||
+ | 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>): | ||
+ | |||
+ | |||
+ | |||
+ | <cmath>DQ^2+PQ^2=PD^2</cmath> | ||
+ | |||
+ | |||
+ | <cmath>\left(\frac{1}{2}\right)^2+PQ^2 = \left(\frac{\sqrt{3}}{2}\right)^2</cmath> | ||
+ | |||
+ | |||
+ | <cmath>PQ^2=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}</cmath> | ||
+ | |||
+ | |||
+ | <cmath>PQ=\frac{\sqrt{2}}{2}\Rightarrow \boxed{\textbf{C}}</cmath> | ||
+ | |||
+ | |||
+ | -WannabeCharmander | ||
== See also == | == See also == |
Latest revision as of 22:35, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.