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 === | ||
− | |||
− | + | <asy> | |
+ | import three; | ||
+ | pointpen = black; | ||
+ | pathpen = black+linewidth(0.7); | ||
+ | pen small = fontsize(9); | ||
+ | currentprojection = perspective(20,-20,12); | ||
+ | triple O=(0,0,0); | ||
+ | triple A=(0, 399^(0.5), 0); | ||
+ | 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 <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> | |
+ | size(280); | ||
+ | import three; | ||
+ | pointpen = black; | ||
+ | pathpen = black+linewidth(0.7); | ||
+ | pen small = fontsize(9); | ||
+ | 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); | ||
+ | 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>. | |
− | <cmath>\begin{ | + | From here we just need to perform some brutish calculations. Using the formula <math>A = 18\sqrt{133} = \frac{abc}{4R}</math> (where <math>R</math> is the [[circumradius]]), we find <math>R = \frac{12\sqrt{3} \cdot (13\sqrt{3})^2}{4\cdot 18\sqrt{133}} = \frac{13^2\sqrt{3}}{2\sqrt{133}}</math> (there are slightly [[Law of Sines|simpler ways]] to calculate <math>R</math> since we have an isosceles triangle). By the Pythagorean Theorem, |
− | &= | + | |
− | &= | + | <cmath> |
− | + | \begin{align*}h^2 &= PA^2 - R^2 \\ | |
− | h &= | + | &= \left(\frac{\sqrt{939}}{2}\right)^2 - \left(\frac{13^2\sqrt{3}}{2\sqrt{133}}\right)^2\\ |
− | \end{ | + | &= \frac{939 \cdot 133 - 13^4 \cdot 3}{4 \cdot 133} = \frac{13068 \cdot 3}{4 \cdot 133} = \frac{99^2}{133}\\ |
+ | h &= \frac{99}{\sqrt{133}} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
Finally, we substitute <math>h</math> into the volume equation to find <math>V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}</math>. | Finally, we substitute <math>h</math> into the volume equation to find <math>V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}</math>. | ||
=== Solution 2 === | === Solution 2 === | ||
− | Let <math>\triangle{ABC}</math> (or the triangle with sides <math>12\sqrt {3}</math>, <math>13\sqrt {3}</math>, <math>13\sqrt {3}</math>) be the base of our tetrahedron. We set points <math> | + | Let <math>\triangle{ABC}</math> (or the triangle with sides <math>12\sqrt {3}</math>, <math>13\sqrt {3}</math>, <math>13\sqrt {3}</math>) be the base of our tetrahedron. We set points <math>C</math> and <math>D</math> as <math>(6\sqrt {3}, 0, 0)</math> and <math>( - 6\sqrt {3}, 0, 0)</math>, respectively. Using Pythagoras, we find <math>A</math> as <math>(0, \sqrt {399}, 0)</math>. We know that the [[vertex]] of the tetrahedron (<math>P</math>) has to be of the form <math>(x, y, z)</math>, where <math>z</math> is the [[altitude]] of the tetrahedron. Since the distance from <math>P</math> to points <math>A</math>, <math>B</math>, and <math>C</math> is <math>\frac {\sqrt {939}}{2}</math>, we can write three equations using the [[distance formula]]: |
− | <cmath>\begin{ | + | <cmath> |
− | x^{2} + (y - \sqrt {399})^{2} + z^{2} &= | + | \begin{align*} |
− | (x - 6\sqrt {3})^{2} + y^{2} + z^{2} &= | + | x^{2} + (y - \sqrt {399})^{2} + z^{2} &= \frac {939}{4}\\ |
− | (x + 6\sqrt {3})^{2} + y^{2} + z^{2} &= | + | (x - 6\sqrt {3})^{2} + y^{2} + z^{2} &= \frac {939}{4}\\ |
− | \end{ | + | (x + 6\sqrt {3})^{2} + y^{2} + z^{2} &= \frac {939}{4} |
+ | \end{align*} | ||
+ | </cmath> | ||
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>. | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:07, 17 September 2016
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.
Solution
Solution 1
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 .
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.