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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>.
  
 
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>.
  
−
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>.
+
<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>.  
 
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>
+
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 PCD)</cmath>
+
<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>
 +
 
 +
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>
 +
 
  
−
<math>\boxed{\textbf{C}}</math>
+
-WannabeCharmander
  
 
== See also ==
 
== See also ==

Latest revision as of 22:35, 21 June 2018

Problem 23

The edges of a regular tetrahedron with vertices $A ,~ B,~ C$, and $D$ each have length one. Find the least possible distance between a pair of points $P$ and $Q$, where $P$ is on edge $AB$ and $Q$ is on edge $CD$.


[asy] size(150); import patterns; 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; aux=midpoint(AA--BB); draw(BB--DD); P=midpoint(AA--aux); aux=midpoint(CC--DD); Q=midpoint(CC--aux); draw(AA--CC,dashed); dot(P); dot(Q); fill(DD--BB--CC--cycle,pattern("hatch")); 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[/asy]


$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{3}{4}\qquad \textbf{(C) }\frac{\sqrt{2}}{2}\qquad \textbf{(D) }\frac{\sqrt{3}}{2}\qquad \textbf{(E) }\frac{\sqrt{3}}{3}$

Solution 1

Note that the distance $PQ$ will be minimized when $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$.

To find this distance, consider triangle $\triangle PCQ$. $Q$ is the midpoint of $CD$, so $CQ=\frac{1}{2}$. Additionally, since $CP$ is the altitude of equilateral $\triangle ABC$, $CP=\frac{\sqrt{3}}{2}$.

[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 $\cos(\angle PCQ)$ in order to find $PQ$ by the Law of Cosines. To do so, drop down $D$ onto $\triangle ABC$ to get the point $D^\prime$.

$\angle PCD$ is congruent to $\angle D^\prime CD$, since $P$, $D^\prime$, and $C$ are collinear. Therefore, we can just find $\cos(\angle D^\prime CD)$.

Note that $\triangle CD^\prime D$ is a right triangle with $\angle CD^\prime D$ 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, $CD=1$.

Note that $D^\prime$ is the centroid of equilateral $\triangle ABC$. Additionally, since $\triangle ABC$ is equilateral, $D^\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\prime$ is $\frac{2}{3}$ of the altitude of $\triangle ABC$. Therefore, $CD^\prime=\frac{\sqrt{3}}{3}$.

Since $\cos(\angle D^\prime CD)=\cos(\angle PCQ)=\frac{CD^\prime}{CD}$, $\cos(\angle PCQ)=\frac{\frac{\sqrt{3}}{3}}{1}=\frac{\sqrt{3}}{3}$

\[PQ^2=CP^2+CQ^2-2(CP)(CQ)\cos(\angle PCQ)\] \[PQ^2=\frac{3}{4}+\frac{1}{4}-2\left(\frac{\sqrt{3}}{4}\right)\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{3}\right)\] Simplifying, $PQ^2=\frac{1}{2}$. Therefore, $PQ=\frac{\sqrt{2}}{2}\Rightarrow$ $\boxed{\textbf{C}}$

Solution by treetor10145

Solution 2 (less overkill)

Notice, like above said, that $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$.


To find the length of $PQ$, first draw in lines $CP$ and $DP$. Notice that $DP$ is an altitude of $\triangle ADP$. We find that $\angle{DAP} = 60 ^{\circ}$ (since $\triangle ABD$ is equilateral), and $AD=\frac{1}{2}$. Use the properties of 30-60-90 triangles to get $DP=\frac{\sqrt{3}}{2}$. Since $CP$ is an altitude of a congruent equilateral triangle, $CP=DP=\frac{\sqrt{3}}{2}$.


Notice that $\triangle CDP$ is isosceles with $CP=DP$. Also, since $Q$ is the midpoint of base $CD$, we can conclude that $PQ$ is an altitude. We can use Pythagorean theorem to get the following (taking into consideration $DQ=\frac{1}{2}$):


\[DQ^2+PQ^2=PD^2\]


\[\left(\frac{1}{2}\right)^2+PQ^2 = \left(\frac{\sqrt{3}}{2}\right)^2\]


\[PQ^2=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\]


\[PQ=\frac{\sqrt{2}}{2}\Rightarrow \boxed{\textbf{C}}\]


-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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