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

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Assume points <math>S</math>, <math>T</math>, and <math>U</math> are on edges <math>AB</math>, <math>AC</math>, and <math>BC</math> respectively such that <math>ES \perp AB</math>, <math>ET \perp AC</math>, and <math>EU \perp BC</math>.
 
Assume points <math>S</math>, <math>T</math>, and <math>U</math> are on edges <math>AB</math>, <math>AC</math>, and <math>BC</math> respectively such that <math>ES \perp AB</math>, <math>ET \perp AC</math>, and <math>EU \perp BC</math>.
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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>.
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It remains to find <math>\frac{EP}{ES}</math>, or equivalently, <math>\sin(\angle DSE)</math>.
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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>.
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== Solution 2(Cheesy) ==
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Continue to assume points <math>P</math>, <math>Q</math>, and <math>R</math> are on faces <math>ABD</math>, <math>ACD</math>, and <math>BCD</math> respectively such that <math>EP \perp ABD</math>, <math>EQ \perp ACD</math>, and <math>ER \perp BCD</math> and assume points <math>S</math>, <math>T</math>, and <math>U</math> are on edges <math>AB</math>, <math>AC</math>, and <math>BC</math> respectively such that <math>ES \perp AB</math>, <math>ET \perp AC</math>, and <math>EU \perp BC</math>.
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Now, because they never specify where E has to be or how long the tetrahedron's sides must be, WLOG, assume E is the centroid/incenter/circumcenter/orthocenter of triangle ABC and the side length of the tetrahedron is <math>2</math> Note that the inradius of ABC is the same as <math>ES=ET=EU.</math> Then the inradius of <math>\triangle\text{ABC}</math> are <math>\frac{\sqrt3}{6}\cdot 2=\frac{\sqrt3}{3}.</math>
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Next, take cross section <math>\triangle\text{SDC}.</math> Since <math>ES</math> is an inradius, <math>ES=\frac{\sqrt3}{3}.</math> Since <math>SD</math> is an altitude of <math>\triangle\text{DAB}</math> and <math>SC</math> is an altitude of <math>\triangle\text{ABC}</math>, <math>SC=SD=\frac{\sqrt3}{2}\cdot 2=\sqrt3.</math> Thus, by the Pythagorean theorem, <math>ED=\sqrt{SD^2-ES^2}=\sqrt{\sqrt{3}^2-\frac{\sqrt{3}^2}{3^2}}=\sqrt{3-\frac13}=\sqrt{\frac83}=\frac{2\sqrt6}{3}.</math>
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Then, because <math>\angle{SED}=\angle{SQE}=90^\circ</math> and <math>\angle{ESD}=\angle{QSE},</math> <math>\triangle{SED}~\triangle{SQE}</math> by AA Similarity. Then, since <math>ES=\frac{SD}3=\frac{\sqrt3}3,</math> <math>EQ=\frac{ED}3=\frac{\frac{2\sqrt6}3}{3}=\frac{2\sqrt6}9.</math>
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Then, <math>s=3\cdot EQ=3\cdot\frac{2\sqrt6}9=\frac{2\sqrt6}3</math> and <math>S=3\cdot\text{inradius}=3\cdot\frac{\sqrt3}3=\sqrt3.</math> Thus, <math>\frac{s}{S}=\frac{\frac{2\sqrt6}3}{\sqrt3}=\frac{2\sqrt2}3=\boxed{\text{(B)}}.</math>
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~~AndrewZhong2012~~
  
 
== 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 10:39, 6 August 2025

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

Solution 2(Cheesy)

Continue to assume points $P$, $Q$, and $R$ are on faces $ABD$, $ACD$, and $BCD$ respectively such that $EP \perp ABD$, $EQ \perp ACD$, and $ER \perp BCD$ and assume points $S$, $T$, and $U$ are on edges $AB$, $AC$, and $BC$ respectively such that $ES \perp AB$, $ET \perp AC$, and $EU \perp BC$.

Now, because they never specify where E has to be or how long the tetrahedron's sides must be, WLOG, assume E is the centroid/incenter/circumcenter/orthocenter of triangle ABC and the side length of the tetrahedron is $2$ Note that the inradius of ABC is the same as $ES=ET=EU.$ Then the inradius of $\triangle\text{ABC}$ are $\frac{\sqrt3}{6}\cdot 2=\frac{\sqrt3}{3}.$

Next, take cross section $\triangle\text{SDC}.$ Since $ES$ is an inradius, $ES=\frac{\sqrt3}{3}.$ Since $SD$ is an altitude of $\triangle\text{DAB}$ and $SC$ is an altitude of $\triangle\text{ABC}$, $SC=SD=\frac{\sqrt3}{2}\cdot 2=\sqrt3.$ Thus, by the Pythagorean theorem, $ED=\sqrt{SD^2-ES^2}=\sqrt{\sqrt{3}^2-\frac{\sqrt{3}^2}{3^2}}=\sqrt{3-\frac13}=\sqrt{\frac83}=\frac{2\sqrt6}{3}.$

Then, because $\angle{SED}=\angle{SQE}=90^\circ$ and $\angle{ESD}=\angle{QSE},$ $\triangle{SED}~\triangle{SQE}$ by AA Similarity. Then, since $ES=\frac{SD}3=\frac{\sqrt3}3,$ $EQ=\frac{ED}3=\frac{\frac{2\sqrt6}3}{3}=\frac{2\sqrt6}9.$

Then, $s=3\cdot EQ=3\cdot\frac{2\sqrt6}9=\frac{2\sqrt6}3$ and $S=3\cdot\text{inradius}=3\cdot\frac{\sqrt3}3=\sqrt3.$ Thus, $\frac{s}{S}=\frac{\frac{2\sqrt6}3}{\sqrt3}=\frac{2\sqrt2}3=\boxed{\text{(B)}}.$

~~AndrewZhong2012~~

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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