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

Latest revision as of 20:34, 27 January 2023

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

@@ Line 2: / Line 2: @@
 Consider the paper triangle whose vertices are <math>(0,0), (34,0),</math> and <math>(16,24).</math>  The vertices of its midpoint triangle are the [[midpoint]]s 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 ==
+== Solution 1 ==
 <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 -->
-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>.
+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. The reason why we find the orthocenter is due to the fact that once we fold the tetrahedron, finding the height requires perpendicular segments that are parallel to the large sides of the large triangle.
+<b>Lemma:</b> The point <math>O</math> is the orthocenter of <math>\triangle ABC</math>.
+<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.
@@ Line 14: / Line 17: @@
 The area of the base is <math>102</math>, so the volume is <math>\frac{102*12}{3}=\boxed{408}</math>.
-==Alternate Solution 1==
+== Solution 2 ==
 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>.
@@ Line 20: / Line 23: @@
 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>.
-==Alternate Solution 2==
+== Solution 3 ==
 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)
@@ Line 38: / Line 41: @@
 Solution by D. Adrian Tanner
+== Solution 4 ==
+Let <math>A = (0,0), B = (16, 24), C = (34,0).</math> Then define <math>D,E,F</math> as the midpoints of <math>BC, AC, AB</math>. By Pythagorean theorem, <math>EF = \frac{1}{2} BC = 15, DE = \frac{1}{2}AB = 4 \sqrt{13}, DF = \frac{1}{2} AC = 17.</math> Then let <math>P</math> be the point in space which is the vertex of the tetrahedron with base <math>DEF</math>.
+Note that <math>\triangle DEP \cong \triangle EDF</math>. Create point <math>F'</math> on the plane of <math>DEF</math> such that <math>\triangle DEP \cong \triangle DEF'</math> (i.e by reflecting <math>F</math> over the perpendicular bisector of <math>DE</math>). Project <math>F, P</math> onto <math>DE</math> as <math>X, Y</math>. Note by the definition of <math>F'</math> then <math>\angle PYF'</math> is the dihedral angle between planes <math>DEP, DEF</math>.
+Now see that by Heron's, <cmath>[DEP] = [DEF] = \sqrt{(16 + 2 \sqrt{13})(16 - 2 \sqrt{13})(1 + 2 \sqrt{13})(-1 + 2 \sqrt{13})} = 102.</cmath>
+So <math>PY</math>, the hypotenuse <math>DEP</math> has length <math>\frac{102 \cdot 2}{4 \sqrt{13}} = \frac{51}{\sqrt{13}}</math>. Similarly <math>F'Y = \frac{51}{\sqrt{13}}.</math> Further from Pythagoras <math>DY = \sqrt{DP^2 - PY^2} = \frac{18}{\sqrt{13}}.</math> Symmetrically <math>EX = \frac{18}{\sqrt{13}}.</math> Therefore <math>XY = DE - DY - EX = \frac{16}{\sqrt{13}}.</math>
+By Law of Cosines on <math>\triangle PYF'</math>,
+<cmath>\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*}.</cmath>
+Therefore the altitude of the tetrahedron from vertex <math>P</math> to base <math>DEF</math> is <math>PY \sin{\angle PYF'} = \frac{51}{\sqrt{13}} \frac{4 \sqrt{13}}{17} = 12.</math> So the area is <math>\frac{1}{3}bh = \frac{1}{3} 12 \cdot 102 = \boxed{408}.</math>
+~ Aaryabhatta1
 == See also ==

1999 AIME (Problems • Answer Key • Resources)
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