1999 AIME Problems/Problem 15

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 1

$[asy]defaultpen(fontsize(9)+linewidth(0.63)); pair A=(0,0), B=(16,24), C=(34,0), P=(8,12), Q=(25,12), R=(17,0); draw(A--B--C--A);draw(P--Q--R--P); draw(A--foot(A,B,C));draw(B--foot(B,A,C));draw(C--foot(C,A,B)); label("$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 $\frac{102*12}{3}=\boxed{408}$ .

Solution 2

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 $34 \times 12 = \boxed{408}$ .

Solution 3

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

Solution 4

Let $A = (0,0), B = (16, 24), C = (34,0).$ Then define $D,E,F$ as the midpoints of $BC, AC, AB$ . By Pythagorean theorem, $EF = \frac{1}{2} BC = 15, DE = \frac{1}{2}AB = 4 \sqrt{13}, DF = \frac{1}{2} AC = 17.$ Then let $P$ be the point in space which is the vertex of the tetrahedron with base $DEF$ .

Note that $\triangle DEP \cong \triangle EDF$ . Create point $F'$ on the plane of $DEF$ such that $\triangle DEP \cong \triangle DEF'$ (i.e by reflecting $F$ over the perpendicular bisector of $DE$ ). Project $F, P$ onto $DE$ as $X, Y$ . Note by the definition of $F'$ then $\angle PYF'$ is the dihedral angle between planes $DEP, DEF$ .

Now see that by Heron's, $[DEP] = [DEF] = \sqrt{(16 + 2 \sqrt{13})(16 - 2 \sqrt{13})(1 + 2 \sqrt{13})(-1 + 2 \sqrt{13})} = 102.$ So $PY$ , the hypotenuse $DEP$ has length $\frac{102 \cdot 2}{4 \sqrt{13}} = \frac{51}{\sqrt{13}}$ . Similarly $F'Y = \frac{51}{\sqrt{13}}.$ Further from Pythagoras $DY = \sqrt{DP^2 - PY^2} = \frac{18}{\sqrt{13}}.$ Symmetrically $EX = \frac{18}{\sqrt{13}}.$ Therefore $XY = DE - DY - EX = \frac{16}{\sqrt{13}}.$

By Law of Cosines on $\triangle PYF'$ , $\begin{align*} PF'^2 &= PY^2 + F'Y^2 - 2 \cdot PY \cdot F'Y \cos{\angle PYF'} \\ PF^2 - XY^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'} \\ (4\sqrt{13})^2 - (\frac{16}{\sqrt{13}})^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'} \\ \cos{\angle PYF'} &= \frac{9}{17} \\ \sin{\angle PYF'} &= \frac{4 \sqrt{13}}{17}. \end{align*}.$

Therefore the altitude of the tetrahedron from vertex $P$ to base $DEF$ is $PY \sin{\angle PYF'} = \frac{51}{\sqrt{13}} \frac{4 \sqrt{13}}{17} = 12.$ So the area is $\frac{1}{3}bh = \frac{1}{3} 12 \cdot 102 = \boxed{408}.$

~ Aaryabhatta1

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