Difference between revisions of "1990 AIME Problems/Problem 14"
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The [[rectangle]] <math>ABCD^{}_{}</math> below has dimensions <math>AB^{}_{} = 12 \sqrt{3}</math> and <math>BC^{}_{} = 13 \sqrt{3}</math>. [[Diagonal]]s <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>P^{}_{}</math>. If triangle <math>ABP^{}_{}</math> is cut out and removed, edges <math>\overline{AP}</math> and <math>\overline{BP}</math> are joined, and the figure is then creased along segments <math>\overline{CP}</math> and <math>\overline{DP}</math>, we obtain a [[triangular pyramid]], all four of whose faces are [[isosceles triangle]]s. Find the volume of this pyramid. | The [[rectangle]] <math>ABCD^{}_{}</math> below has dimensions <math>AB^{}_{} = 12 \sqrt{3}</math> and <math>BC^{}_{} = 13 \sqrt{3}</math>. [[Diagonal]]s <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>P^{}_{}</math>. If triangle <math>ABP^{}_{}</math> is cut out and removed, edges <math>\overline{AP}</math> and <math>\overline{BP}</math> are joined, and the figure is then creased along segments <math>\overline{CP}</math> and <math>\overline{DP}</math>, we obtain a [[triangular pyramid]], all four of whose faces are [[isosceles triangle]]s. Find the volume of this pyramid. | ||
− | + | <asy> | |
+ | pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6); | ||
+ | draw(B--C--P--D--C^^D--A); | ||
+ | filldraw(A--P--B--cycle, gray, black); | ||
+ | label("$A$", A, SE); | ||
+ | label("$B$", B, NE); | ||
+ | label("$C$", C, NW); | ||
+ | label("$D$", D, SW); | ||
+ | label("$P$", P, N); | ||
+ | label("$13\sqrt{3}$", A--D, S); | ||
+ | label("$12\sqrt{3}$", A--B, E);</asy> | ||
__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
− | === Solution 1 === | + | === Solution 1(Synthetic) === |
− | + | ||
<asy> | <asy> | ||
− | import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); | + | import three; |
+ | pointpen = black; | ||
+ | pathpen = black+linewidth(0.7); | ||
+ | pen small = fontsize(9); | ||
currentprojection = perspective(20,-20,12); | currentprojection = perspective(20,-20,12); | ||
− | triple O=(0,0,0) | + | triple O=(0,0,0); |
− | pair | + | triple A=(0, 399^(0.5), 0); |
− | triple P=( | + | triple D=(108^(0.5), 0, 0); |
− | + | triple C=(-108^(0.5), 0, 0); | |
− | + | triple Pa; | |
− | + | pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); | |
− | + | triple P=(Ci.x, Ci.y, (99/133)^.5); | |
+ | Pa=(P.x,P.y,0); | ||
+ | draw(P--Pa--A); | ||
+ | draw(C--Pa--D); | ||
+ | draw((C+D)/2--A--C--D--P--C--P--A--D); | ||
+ | label("A", A, NE); | ||
+ | label("P", P, N); | ||
+ | label("C", C); | ||
+ | label("D", D, S); | ||
+ | label("$13\sqrt{3}$", (A+D)/2, E, small); | ||
+ | label("$13\sqrt{3}$", (A+C)/2, N, small); | ||
+ | label("$12\sqrt{3}$", (C+D)/2, SW, small); | ||
+ | </asy> | ||
− | Our triangular pyramid has base | + | Our triangular pyramid has base <math>12\sqrt{3} - 13\sqrt{3} - 13\sqrt{3} \triangle</math>. The area of this isosceles triangle is easy to find by <math>[ACD] = \frac{1}{2}bh</math>, where we can find <math>h_{ACD}</math> to be <math>\sqrt{399}</math> by the [[Pythagorean Theorem]]. Thus <math>A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}</math>. |
<asy> | <asy> | ||
− | |||
size(280); | size(280); | ||
− | import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); | + | import three; |
− | real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); | + | pointpen = black; |
− | triple O=(0,0,0) | + | pathpen = black+linewidth(0.7); |
− | pair | + | pen small = fontsize(9); |
− | triple P=( | + | real h=169/2*(3/133)^.5; |
− | + | currentprojection = perspective(20,-20,12); | |
− | + | triple O=(0,0,0); | |
− | + | triple A=(0,399^.5,0); | |
− | + | triple D=(108^.5,0,0); | |
− | + | triple C=(-108^.5,0,0); | |
− | </asy | + | pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); |
+ | triple P=(Ci.x, Ci.y, 99/133^.5); | ||
+ | triple Pa=(P.x,P.y,0); | ||
+ | draw(A--C--D--P--C--P--A--D); | ||
+ | draw(P--Pa--A); | ||
+ | draw(C--Pa--D); | ||
+ | draw(circle(Pa, h)); | ||
+ | label("A", A, NE); | ||
+ | label("C", C, NW); | ||
+ | label("D", D, S); | ||
+ | label("P",P , N); | ||
+ | label("P$'$", Pa, SW); | ||
+ | label("$13\sqrt{3}$", (A+D)/2, E, small); | ||
+ | label("$13\sqrt{3}$", (A+C)/2, NW, small); | ||
+ | label("$12\sqrt{3}$", (C+D)/2, SW, small); | ||
+ | label("h", (P + Pa)/2, W); | ||
+ | label("$\frac{\sqrt{939}}{2}$", (C+P)/2 ,NW); | ||
+ | </asy> <!-- Asymptote replacement for Image:1990_AIME-14c.png by azjps --> | ||
To find the volume, we want to use the equation <math>\frac 13Bh = 6\sqrt{133}h</math>, so we need to find the height of the [[tetrahedron]]. By the Pythagorean Theorem, <math>AP = CP = DP = \frac{\sqrt{939}}{2}</math>. If we let <math>P</math> be the center of a [[sphere]] with radius <math>\frac{\sqrt{939}}{2}</math>, then <math>A,C,D</math> lie on the sphere. The cross section of the sphere that contains <math>A,C,D</math> is a circle, and the center of that circle is the foot of the [[perpendicular]] from the center of the sphere. Hence the foot of the height we want to find occurs at the [[circumcenter]] of <math>\triangle ACD</math>. | To find the volume, we want to use the equation <math>\frac 13Bh = 6\sqrt{133}h</math>, so we need to find the height of the [[tetrahedron]]. By the Pythagorean Theorem, <math>AP = CP = DP = \frac{\sqrt{939}}{2}</math>. If we let <math>P</math> be the center of a [[sphere]] with radius <math>\frac{\sqrt{939}}{2}</math>, then <math>A,C,D</math> lie on the sphere. The cross section of the sphere that contains <math>A,C,D</math> is a circle, and the center of that circle is the foot of the [[perpendicular]] from the center of the sphere. Hence the foot of the height we want to find occurs at the [[circumcenter]] of <math>\triangle ACD</math>. | ||
Line 63: | Line 104: | ||
Subtracting the last two equations, we get <math>x = 0</math>. Solving for <math>y,z</math> with a bit of effort, we eventually get <math>x = 0</math>, <math>y = \frac {291}{2\sqrt {399}}</math>, <math>z = \frac {99}{\sqrt {133}}</math>. | Subtracting the last two equations, we get <math>x = 0</math>. Solving for <math>y,z</math> with a bit of effort, we eventually get <math>x = 0</math>, <math>y = \frac {291}{2\sqrt {399}}</math>, <math>z = \frac {99}{\sqrt {133}}</math>. | ||
Since the area of a triangle is <math>\frac {1}{2}\cdot bh</math>, we have the base area as <math>18\sqrt {133}</math>. Thus, the volume is <math>V = \frac {1}{3}\cdot18\sqrt {133}\cdot\frac {99}{\sqrt {133}} = 6\cdot99 = 594</math>. | Since the area of a triangle is <math>\frac {1}{2}\cdot bh</math>, we have the base area as <math>18\sqrt {133}</math>. Thus, the volume is <math>V = \frac {1}{3}\cdot18\sqrt {133}\cdot\frac {99}{\sqrt {133}} = 6\cdot99 = 594</math>. | ||
+ | |||
+ | === Solution 3 (Law of Cosines) === | ||
+ | |||
+ | Let <math>X</math> be the apex of the pyramid and <math>M</math> be the midpoint of <math>\overline{CD}</math>. We find the side lengths of <math>\triangle XMP</math>. | ||
+ | |||
+ | <math>MP = \frac{13\sqrt3}{2}</math>. <math>PX</math> is half of <math>AC</math>, which is <math>\frac{\sqrt{3\cdot13^2+3\cdot12^2}}{2} = \frac{\sqrt{939}}{2}</math>. To find <math>MX</math>, consider right triangle <math>XMD</math>; since <math>XD=13\sqrt3</math> and <math>MD=6\sqrt3</math>, we have <math>MX=\sqrt{3\cdot13^2-3\cdot6^2} = \sqrt{399}</math>. | ||
+ | |||
+ | Let <math>\theta=\angle XPM</math>. For calculating trig, let us double all sides of <math>\triangle XMP</math>. By Law of Cosines, <math>\cos\theta = \frac{3\cdot13^2+939-4\cdot399}{2\cdot13\cdot3\sqrt{313}}=\frac{-150}{6\cdot13\sqrt{313}}=-\frac{25}{13\sqrt{313}}</math>. | ||
+ | |||
+ | Hence, <cmath>\sin\theta = \sqrt{1-\cos^2\theta}</cmath> | ||
+ | <cmath>=\sqrt{1-\frac{625}{13^2\cdot313}}</cmath> | ||
+ | <cmath>=\frac{\sqrt{13^2\cdot313-625}}{13\sqrt{313}}</cmath> | ||
+ | <cmath>=\frac{\sqrt{3\cdot132^2}}{13\sqrt{313}}</cmath> | ||
+ | <cmath>=\frac{132\sqrt3}{13\sqrt{313}}</cmath> | ||
+ | Thus, the height of the pyramid is <math>PX\sin\theta=\frac{\sqrt{939}}{2}\cdot\frac{132\sqrt3}{13\sqrt{313}}=\frac{66\cdot3}{13}</math>. Since <math>[CPD]=3\sqrt{3}\cdot13\sqrt{3}=9\cdot13</math>, the volume of the pyramid is <math>\frac13 \cdot 9\cdot13\cdot \frac{66\cdot3}{13}=\boxed{594}</math>. | ||
== See also == | == See also == |
Latest revision as of 08:16, 29 July 2023
Problem
The rectangle below has dimensions
and
. Diagonals
and
intersect at
. If triangle
is cut out and removed, edges
and
are joined, and the figure is then creased along segments
and
, we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid.
Contents
Solution
Solution 1(Synthetic)
Our triangular pyramid has base . The area of this isosceles triangle is easy to find by
, where we can find
to be
by the Pythagorean Theorem. Thus
.
To find the volume, we want to use the equation , so we need to find the height of the tetrahedron. By the Pythagorean Theorem,
. If we let
be the center of a sphere with radius
, then
lie on the sphere. The cross section of the sphere that contains
is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of
.
From here we just need to perform some brutish calculations. Using the formula (where
is the circumradius), we find
(there are slightly simpler ways to calculate
since we have an isosceles triangle). By the Pythagorean Theorem,
Finally, we substitute into the volume equation to find
.
Solution 2
Let (or the triangle with sides
,
,
) be the base of our tetrahedron. We set points
and
as
and
, respectively. Using Pythagoras, we find
as
. We know that the vertex of the tetrahedron (
) has to be of the form
, where
is the altitude of the tetrahedron. Since the distance from
to points
,
, and
is
, we can write three equations using the distance formula:
Subtracting the last two equations, we get . Solving for
with a bit of effort, we eventually get
,
,
.
Since the area of a triangle is
, we have the base area as
. Thus, the volume is
.
Solution 3 (Law of Cosines)
Let be the apex of the pyramid and
be the midpoint of
. We find the side lengths of
.
.
is half of
, which is
. To find
, consider right triangle
; since
and
, we have
.
Let . For calculating trig, let us double all sides of
. By Law of Cosines,
.
Hence,
Thus, the height of the pyramid is
. Since
, the volume of the pyramid is
.
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.