Difference between revisions of "1999 AIME Problems/Problem 15"

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<center><asy>defaultpen(fontsize(9)+linewidth(0.63)); pair A=(0,0), B=(16,24), C=(34,0), P=(8,12), Q=(25,12), R=(17,0); draw(A--B--C--A);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,SE); label("\(D\)",foot(A,B,C),NE);label("\(E\)",foot(B,A,C),SW);label("\(F\)",foot(C,A,B),NW);label("\(P\)",P,NW);label("\(Q\)",Q,NE);label("\(R\)",R,SE);</asy><asy>import three; defaultpen(linewidth(0.6));
 
<center><asy>defaultpen(fontsize(9)+linewidth(0.63)); pair A=(0,0), B=(16,24), C=(34,0), P=(8,12), Q=(25,12), R=(17,0); draw(A--B--C--A);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,SE); label("\(D\)",foot(A,B,C),NE);label("\(E\)",foot(B,A,C),SW);label("\(F\)",foot(C,A,B),NW);label("\(P\)",P,NW);label("\(Q\)",Q,NE);label("\(R\)",R,SE);</asy><asy>import three; defaultpen(linewidth(0.6));
 
currentprojection=orthographic(1/2,-1,1/2); triple A=(0,0,0), B=(16,24,0), C=(34,0,0), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0)); </asy></center><!-- Asymptote renderings of Image:AIME_1999_Solution_15_1.png, Image:AIME_1999_Solution_15_2.png, by Minsoens -->
 
currentprojection=orthographic(1/2,-1,1/2); triple A=(0,0,0), B=(16,24,0), C=(34,0,0), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0)); </asy></center><!-- Asymptote renderings of Image:AIME_1999_Solution_15_1.png, Image:AIME_1999_Solution_15_2.png, by Minsoens -->
Let <math>D</math>, <math>E</math>, <math>F</math> be the feet of the altitudes to sides <math>BC</math>, <math>CA</math>, <math>AB</math>, respectively, of <math>\triangle ABC</math>.
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As shown in the image above, let <math>D</math>, <math>E</math>, and <math>F</math> be the midpoints of <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math>, respectively.  Suppose <math>P</math> is the apex of the tetrahedron, and let <math>O</math> be the foot of the altitude from <math>P</math> to <math>\triangle ABC</math>. The crux of this problem is the following lemma.
The base of the [[tetrahedron]] is the [[orthocenter]] <math>O</math> of the large triangle, so we just need to find that, then it's easy from there.
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<b>Lemma:</b> The point <math>O</math> is the orthocenter of <math>\triangle ABC</math>.
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<i>Proof.</i> Observe that <cmath>OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;</cmath> the first equality follows by the Pythagorean Theorem, while the second follows from <math>AF = FP</math> and <math>AE = EP</math>.  Thus, by the Perpendicularity Lemma, <math>AO</math> is perpendicular to <math>FE</math> and hence <math>BC</math>.  Analogously, <math>O</math> lies on the <math>B</math>-altitude and <math>C</math>-altitude of <math>\triangle ABC</math>, and so <math>O</math> is, indeed, the orthocenter of <math>\triangle ABC</math>.
  
 
To find the coordinates of <math>O</math>, we need to find the intersection point of altitudes <math>BE</math> and <math>AD</math>. The equation of <math>BE</math> is simply <math>x=16</math>. <math>AD</math> is [[perpendicular]] to line <math>BC</math>, so the slope of <math>AD</math> is equal to the negative reciprocal of the slope of <math>BC</math>. <math>BC</math> has slope <math>\frac{24-0}{16-34}=-\frac{4}{3}</math>, therefore <math>y=\frac{3}{4} x</math>. These two lines intersect at <math>(16,12)</math>, so that's the base of the height of the tetrahedron.  
 
To find the coordinates of <math>O</math>, we need to find the intersection point of altitudes <math>BE</math> and <math>AD</math>. The equation of <math>BE</math> is simply <math>x=16</math>. <math>AD</math> is [[perpendicular]] to line <math>BC</math>, so the slope of <math>AD</math> is equal to the negative reciprocal of the slope of <math>BC</math>. <math>BC</math> has slope <math>\frac{24-0}{16-34}=-\frac{4}{3}</math>, therefore <math>y=\frac{3}{4} x</math>. These two lines intersect at <math>(16,12)</math>, so that's the base of the height of the tetrahedron.  
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The area of the base is <math>102</math>, so the volume is <math>\frac{102*12}{3}=\boxed{408}</math>.
 
The area of the base is <math>102</math>, so the volume is <math>\frac{102*12}{3}=\boxed{408}</math>.
  
==Alternate Solution==
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==Alternate Solution 1==
  
 
Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates <math>(17, 0, 0)</math>, <math>(8, 12, 0)</math>, and <math>(25, 12, 0)</math>. We can compute the area of this triangle as <math>102</math>. Suppose <math>(x, y, z)</math> are the coordinates of the vertex of the resulting pyramid. Call this point <math>V</math>. Clearly, the height of the pyramid is <math>z</math>. The desired volume is thus <math>\frac{102z}{3} = 34z</math>.  
 
Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates <math>(17, 0, 0)</math>, <math>(8, 12, 0)</math>, and <math>(25, 12, 0)</math>. We can compute the area of this triangle as <math>102</math>. Suppose <math>(x, y, z)</math> are the coordinates of the vertex of the resulting pyramid. Call this point <math>V</math>. Clearly, the height of the pyramid is <math>z</math>. The desired volume is thus <math>\frac{102z}{3} = 34z</math>.  
  
 
We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, <math>VR = RA</math>, <math>VP = PB</math>, and <math>VQ = QC</math>. We then use distance formula to find the distances from <math>V</math> to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding <math>z = 12</math>. The desired volume is thus <math>34 \times 12 = \boxed{408}</math>.
 
We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, <math>VR = RA</math>, <math>VP = PB</math>, and <math>VQ = QC</math>. We then use distance formula to find the distances from <math>V</math> to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding <math>z = 12</math>. The desired volume is thus <math>34 \times 12 = \boxed{408}</math>.
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==Alternate Solution 2==
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The formed tetrahedron has pairwise parallel planar and oppositely equal length (<math>4\sqrt{13},15,17</math>) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces.  Let the edge lengths of the parallelepiped be <math>p,q,r</math> and solve (by Pythagoras)
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<math>p^2+q^2=4^2\cdot{13}</math>
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<math>q^2+r^2=15^2</math>
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<math>r^2+p^2=17^2</math>
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to find that <math>(p^2,q^2,r^2)=(153,136,72)=(3^2\cdot{17},2^3\cdot{17},2^3\cdot{3^2}).</math>
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Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is <math>\tfrac{1}{3}</math> and then the volume is
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<math>\tfrac{1}{3}pqr=\tfrac{1}{3}\sqrt{2^6\cdot{3^4}\cdot{17^2}}=\boxed{408}</math>
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Solution by D. Adrian Tanner
  
 
== See also ==
 
== See also ==

Revision as of 00:46, 20 January 2020

Problem

Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?

Solution

[asy]defaultpen(fontsize(9)+linewidth(0.63)); pair A=(0,0), B=(16,24), C=(34,0), P=(8,12), Q=(25,12), R=(17,0); draw(A--B--C--A);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,SE); label("\(D\)",foot(A,B,C),NE);label("\(E\)",foot(B,A,C),SW);label("\(F\)",foot(C,A,B),NW);label("\(P\)",P,NW);label("\(Q\)",Q,NE);label("\(R\)",R,SE);[/asy][asy]import three; defaultpen(linewidth(0.6)); currentprojection=orthographic(1/2,-1,1/2); triple A=(0,0,0), B=(16,24,0), C=(34,0,0), P=(8,12,0), Q=(25,12,0), R=(17,0,0), S=(16,12,12); draw(A--B--C--A); draw(P--Q--R--P); draw(S--P..S--Q..S--R); draw(S--(16,12,0)); [/asy]

As shown in the image above, let $D$, $E$, and $F$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\triangle ABC$. The crux of this problem is the following lemma.

Lemma: The point $O$ is the orthocenter of $\triangle ABC$.

Proof. Observe that \[OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;\] the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$. Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$. Analogously, $O$ lies on the $B$-altitude and $C$-altitude of $\triangle ABC$, and so $O$ is, indeed, the orthocenter of $\triangle ABC$.

To find the coordinates of $O$, we need to find the intersection point of altitudes $BE$ and $AD$. The equation of $BE$ is simply $x=16$. $AD$ is perpendicular to line $BC$, so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$. $BC$ has slope $\frac{24-0}{16-34}=-\frac{4}{3}$, therefore $y=\frac{3}{4} x$. These two lines intersect at $(16,12)$, so that's the base of the height of the tetrahedron.

Let $S$ be the foot of altitude $BS$ in $\triangle BPQ$. From the Pythagorean Theorem, $h=\sqrt{BS^2-SO^2}$. However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$.

The area of the base is $102$, so the volume is $\frac{102*12}{3}=\boxed{408}$.

Alternate Solution 1

Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates $(17, 0, 0)$, $(8, 12, 0)$, and $(25, 12, 0)$. We can compute the area of this triangle as $102$. Suppose $(x, y, z)$ are the coordinates of the vertex of the resulting pyramid. Call this point $V$. Clearly, the height of the pyramid is $z$. The desired volume is thus $\frac{102z}{3} = 34z$.

We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, $VR = RA$, $VP = PB$, and $VQ = QC$. We then use distance formula to find the distances from $V$ to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding $z = 12$. The desired volume is thus $34 \times 12 = \boxed{408}$.

Alternate Solution 2

The formed tetrahedron has pairwise parallel planar and oppositely equal length ($4\sqrt{13},15,17$) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces. Let the edge lengths of the parallelepiped be $p,q,r$ and solve (by Pythagoras)

$p^2+q^2=4^2\cdot{13}$

$q^2+r^2=15^2$

$r^2+p^2=17^2$

to find that $(p^2,q^2,r^2)=(153,136,72)=(3^2\cdot{17},2^3\cdot{17},2^3\cdot{3^2}).$

Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is $\tfrac{1}{3}$ and then the volume is

$\tfrac{1}{3}pqr=\tfrac{1}{3}\sqrt{2^6\cdot{3^4}\cdot{17^2}}=\boxed{408}$


Solution by D. Adrian Tanner

See also

1999 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
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