# 1979 AHSME Problems/Problem 23

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## 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