# Difference between revisions of "2017 AIME I Problems/Problem 4"

## Problem 4

A pyramid has a triangular base with side lengths $20$, $20$, and $24$. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$. The volume of the pyramid is $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

## Solution

Let the triangular base be $\triangle ABC$, with $\overline {AB} = 24$. We find that the altitude to side $\overline {AB}$ is $16$, so the area of $\triangle ABC$ is $(24*16)/2 = 192$.

Let the fourth vertex of the tetrahedron be $P$, and let the midpoint of $\overline {AB}$ be $M$. Since $P$ is equidistant from $A$, $B$, and $C$, the line through $P$ perpendicular to the plane of $\triangle ABC$ will pass through the circumcenter of $\triangle ABC$, which we will call $O$. Note that $O$ is equidistant from each of $A$, $B$, and $C$. Then,

$$\overline {OM} + \overline {OC} = \overline {CM} = 16$$

Let $\overline {OM} = d$. Equation $(1)$: $$d + \sqrt {d^2 + 144} = 16$$

Squaring both sides, we have

$$d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256$$

$$2d^2 + 2d\sqrt {d^2+144} = 112$$

$$2d(d + \sqrt {d^2+144}) = 112$$

Substituting with equation $(1)$:

$$2d(16) = 112$$

$$d = 7/2$$

We now find that $\sqrt{d^2 + 144} = 25/2$.

Let the distance $\overline {OP} = h$. Using the Pythagorean Theorem on triangle $AOP$, $BOP$, or $COP$ (all three are congruent by SSS):

$$25^2 = h^2 + (25/2)^2$$

$$625 = h^2 + 625/4$$

$$1875/4 = h^2$$

$$25\sqrt {3} / 2 = h$$

Finally, by the formula for volume of a pyramid,

$$V = Bh/3$$

$$V = (192)(25\sqrt{3}/2)/3$$ This simplifies to $V = 800\sqrt {3}$, so $m+n = \boxed {803}$.

## Shortcut

Here is a shortcut for finding the radius $R$ of the circumcenter of $\triangle ABC$.

As before, we find that the foot of the altitude from $P$ lands on the circumcenter of $\triangle ABC$. Let $BC=a$, $AC=b$, and $AB=c$. Then we write the area of $\triangle ABC$ in two ways: $$[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}$$

Plugging in $20$, $20$, and $24$ for $a$, $b$, and $c$ respectively, and solving for $R$, we obtain $R= \frac{25}{2}=OA=OB=OC$.

Then continue as before to use the Pythagorean Theorem on $\triangle AOP$, find $h$, and find the volume of the pyramid.

## Solution 2 (Coordinates)

We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length $24$ is at the origin, or $(0, 0, 0)$. Then, the two other vertices can be $(-12, -16, 0)$ and $(12, -16, 0)$. Let the fourth vertex have coordinates of $(x, y, z)$. We have the following $3$ equations from the distance formula.

$$x^2+y^2+z^2=625$$

$$(x+12)^2+(y+16)^2+z^2=625$$

$$(x-12)^2+(y+16)^2+z^2=625$$

Adding the last two equations and substituting in the first equation, we get that $y=-\frac{25}{2}$. If you drew a good diagram, it should be obvious that $x=0$. Now, solving for $z$, we get that $z=\frac{25\sqrt{3}}{2}$. So, the height of the pyramid is $\frac{25\sqrt{3}}{2}$. The base is equal to the area of the triangle, which is $\frac{1}{2} \cdot 24 \cdot 16 = 192$. The volume is $\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}$. Thus, the answer is $800+3 = \boxed{803}$.

-RootThreeOverTwo

## See Also

 2017 AIME I (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 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.

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