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Difference between revisions of "2002 AMC 12P Problems/Problem 24"

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Consider triangles <math>EPS</math>, <math>EQT</math>, and <math>ERU</math>. 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 <math>\frac{EP}{ES} = \frac{EQ}{ET} = \frac{ER}{EU} = \frac{EP+EQ+ER}{ES+ET+EU} = \frac{s}{S}</math>.
 
Consider triangles <math>EPS</math>, <math>EQT</math>, and <math>ERU</math>. 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 <math>\frac{EP}{ES} = \frac{EQ}{ET} = \frac{ER}{EU} = \frac{EP+EQ+ER}{ES+ET+EU} = \frac{s}{S}</math>.
  
−
It remains to find <math>\frac{EP}{ES}</math>, or equivalently, <math>cos(\angle DSE)</math>.
+
It remains to find <math>\frac{EP}{ES}</math>, or equivalently, <math>\sin(\angle DSE)</math>.
 +
 
 +
We know <math>SE = \frac{1}{3}DS</math> by the centroid property. Therefore, <math>\cos(\angle DSE) = \frac{1}{3}</math>, so <math>\sin(\angle DSE) = \sqrt{1-(\frac{1}{3})^2} = \boxed {\text{(B) }\frac{2 \sqrt{2}}{3}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=23|num-a=25}}
 
{{AMC12 box|year=2002|ab=P|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:04, 10 March 2024

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 $\sin(\angle DSE) = \sqrt{1-(\frac{1}{3})^2} = \boxed {\text{(B) }\frac{2 \sqrt{2}}{3}}$.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AMC 12 Problems and Solutions

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