Difference between revisions of "1999 AIME Problems/Problem 15"
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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? | 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)); | <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 --> | ||
− | + | 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 | + | |
+ | <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. | 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>. | The area of the base is <math>102</math>, so the volume is <math>\frac{102*12}{3}=\boxed{408}</math>. | ||
− | == | + | == 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>. | 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 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>. |
+ | |||
+ | == 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) | ||
+ | |||
+ | <math>p^2+q^2=4^2\cdot{13}</math> | ||
+ | |||
+ | <math>q^2+r^2=15^2</math> | ||
+ | |||
+ | <math>r^2+p^2=17^2</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <math>\tfrac{1}{3}pqr=\tfrac{1}{3}\sqrt{2^6\cdot{3^4}\cdot{17^2}}=\boxed{408}</math> | ||
+ | |||
+ | |||
+ | 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 | ||
+ | |||
+ | == Solution 5 (Formula Abuse) == | ||
+ | |||
+ | The Pyramid is a disphenoid, because opposite sides have the same length. The volume of a disphenoid is given by | ||
+ | |||
+ | <cmath>V = \sqrt{\frac{(l^{2}+m^{2}-n^{2})(l^{2}-m^{2}+n^{2})(-l^{2}+m^{2}+n^{2})}{72}}.</cmath> | ||
+ | |||
+ | Using the Pythagorean theorem, the side lengths of the smaller triangle are \(15\), \(4\sqrt{13}\), and \(17\). Plugging in, we get | ||
+ | |||
+ | <cmath>V = \sqrt{166464} = \boxed{408}.</cmath> | ||
+ | |||
+ | ~~Disphenoid_lover | ||
== See also == | == See also == |
Latest revision as of 15:28, 13 May 2024
Contents
Problem
Consider the paper triangle whose vertices are and
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]](http://latex.artofproblemsolving.com/8/b/c/8bca170a0544db80c5bdf360300276df0efcb5e9.png)
![[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]](http://latex.artofproblemsolving.com/0/3/3/0334c5d9ccf52482e6d919c1ccf99cd120d73464.png)
As shown in the image above, let ,
, and
be the midpoints of
,
, and
, respectively. Suppose
is the apex of the tetrahedron, and let
be the foot of the altitude from
to
. The crux of this problem is the following lemma.
Lemma: The point is the orthocenter of
.
Proof. Observe that the first equality follows by the Pythagorean Theorem, while the second follows from
and
. Thus, by the Perpendicularity Lemma,
is perpendicular to
and hence
. Analogously,
lies on the
-altitude and
-altitude of
, and so
is, indeed, the orthocenter of
.
To find the coordinates of , we need to find the intersection point of altitudes
and
. The equation of
is simply
.
is perpendicular to line
, so the slope of
is equal to the negative reciprocal of the slope of
.
has slope
, therefore
. These two lines intersect at
, so that's the base of the height of the tetrahedron.
Let be the foot of altitude
in
. From the Pythagorean Theorem,
. However, since
and
are, by coincidence, the same point,
and
.
The area of the base is , so the volume is
.
Solution 2
Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates ,
, and
. We can compute the area of this triangle as
. Suppose
are the coordinates of the vertex of the resulting pyramid. Call this point
. Clearly, the height of the pyramid is
. The desired volume is thus
.
We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, ,
, and
. We then use distance formula to find the distances from
to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding
. The desired volume is thus
.
Solution 3
The formed tetrahedron has pairwise parallel planar and oppositely equal length () 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
and solve (by Pythagoras)
to find that
Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is and then the volume is
Solution by D. Adrian Tanner
Solution 4
Let Then define
as the midpoints of
. By Pythagorean theorem,
Then let
be the point in space which is the vertex of the tetrahedron with base
.
Note that . Create point
on the plane of
such that
(i.e by reflecting
over the perpendicular bisector of
). Project
onto
as
. Note by the definition of
then
is the dihedral angle between planes
.
Now see that by Heron's,
So
, the hypotenuse
has length
. Similarly
Further from Pythagoras
Symmetrically
Therefore
By Law of Cosines on ,
Therefore the altitude of the tetrahedron from vertex to base
is
So the area is
~ Aaryabhatta1
Solution 5 (Formula Abuse)
The Pyramid is a disphenoid, because opposite sides have the same length. The volume of a disphenoid is given by
Using the Pythagorean theorem, the side lengths of the smaller triangle are \(15\), \(4\sqrt{13}\), and \(17\). Plugging in, we get
~~Disphenoid_lover
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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