2002 AMC 12P Problems/Problem 24

Problem

Let $ABCD$ be a regular tetrahedron and Let $E$ be a point inside the face $ABC.$ Denote by $s$ the sum of the distances from $E$ to the faces $DAB, DBC, DCA,$ and by $S$ the sum of the distances from $E$ to the edges $AB, BC, CA.$ Then $\frac{s}{S}$ equals

$\text{(A) }\sqrt{2} \qquad \text{(B) }\frac{2 \sqrt{2}}{3} \qquad \text{(C) }\frac{\sqrt{6}}{2} \qquad \text{(D) }2 \qquad \text{(E) }3$

Solution

Assume points $P$ , $Q$ , and $R$ are on faces $ABD$ , $ACD$ , and $BCD$ respectively such that $EP \perp ABD$ , $EQ \perp ACD$ , and $ER \perp BCD$ .

Assume points $S$ , $T$ , and $U$ are on edges $AB$ , $AC$ , and $BC$ respectively such that $ES \perp AB$ , $ET \perp AC$ , and $EU \perp BC$ .

Consider triangles $EPS$ , $EQT$ , and $ERU$ . Each of these triangles have a right angle and an angle equal to the dihedral angle of the tetrahedron, so they are all similar by AA similarity. In particular, we know that $\frac{EP}{ES} = \frac{EQ}{ET} = \frac{ER}{EU} = \frac{EP+EQ+ER}{ES+ET+EU} = \frac{s}{S}$ .

It remains to find $\frac{EP}{ES}$ , or equivalently, $\sin(\angle DSE)$ .

We know $SE = \frac{1}{3}DS$ by the centroid property. Therefore, $\cos(\angle DSE) = \frac{1}{3}$ , so $\sin(\angle DSE) = \sqrt{1-(\frac{1}{3})^2} = \frac{2\sqrt{2}}{3}$ . It follows that $\frac{s}{S} = \boxed{\frac{2\sqrt{2}}{3}$ (Error compiling LaTeX. Unknown error_msg)

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2002 AMC 12P Problems/Problem 24

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