# 2017 AIME I Problems/Problem 4

## Problem

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 $m+n = \boxed {803}$.

NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows :

Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that $\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}$. Now integrate it taking the limits $0$ to $H$

### 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.

### Another Shortcut (Extended Law of Sines)

Take the base $\triangle ABC$, where $AB = BC = 20$ and $AC = 24$. Draw an altitude from $B$ to $AC$ that bisects $AC$ at point $D$. Then the altitude has length $\sqrt{20^2 - 12^2} = \sqrt{16^2} = 16$. Next, let $\angle BCA = \theta$. Then from the right triangle $\triangle BDC$, $\sin \theta = 4/5$. From the extended law of sines, the circumradius is $20 \cdot \dfrac{5}{4} \cdot \dfrac{1}{2} = \dfrac{25}{2}$.

## 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

## Solution 3 (Heron's Formula)

Label the four vertices of the tetrahedron and the midpoint of $\overline {AB}$, and notice that the area of the base of the tetrahedron, $\triangle ABC$, equals $192$, according to Solution 1.

Notice that the altitude of $\triangle CPM$ from $\overline {CM}$ to point $P$ is the height of the tetrahedron. Side $\overline {PM}$ is can be found using the Pythagorean Theorem on $\triangle APM$, giving us $\overline {PM}=\sqrt{481}.$

Using Heron's Formula, the area of $\triangle CPM$ can be written as $$\sqrt{\frac{41+\sqrt{481}}{2}(\frac{41+\sqrt{481}}{2}-16)(\frac{41+\sqrt{481}}{2}-25)(\frac{41+\sqrt{481}}{2}-\sqrt{481})}$$ $$=\frac{\sqrt{(41+\sqrt{481})(9+\sqrt{481})(-9+\sqrt{481})(41-\sqrt{481})}}{4}$$

Notice that both $(41+\sqrt{481})(41-\sqrt{481})$ and $(9+\sqrt{481})(-9+\sqrt{481})$ can be rewritten as differences of squares; thus, the expression can be written as $$\frac{\sqrt{(41^2-481)(481-9^2)}}{4}=\frac{\sqrt{480000}}{4}=100\sqrt{3}.$$

From this, we can determine the height of both $\triangle CPM$ and tetrahedron $ABCP$ to be $\frac{100\sqrt{3}}{8}$; therefore, the volume of the tetrahedron equals $\frac{100\sqrt{3}}{8} \cdot 192=800\sqrt{3}$; thus, $m+n=800+3=\boxed{803}.$

-dzhou100

## Solution 4 (Symmetry)

Notation is shown on diagram. $$AM = MB = c = 12, AC = BC = b = 20,$$ $$DA = DB = DC = a = 25.$$ $$CM = x + y = \sqrt{b^2-c^2} = 16,$$ $$x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,$$ $$x – y = \frac{x^2 – y^2}{x+y} = \frac {c^2} {16} = 9,$$ $$x = \frac {16 + 9}{2} = \frac {a}{2},$$ $$h = \sqrt{a^2 -\frac{ a^2}{4}} = a \frac {\sqrt{3}}{2},$$ $$V = \frac{h\cdot CM \cdot c}{3}= \frac{16\cdot 25 \sqrt{3} \cdot 12}{3} = 800 \sqrt{3} \implies \boxed {803}.$$ vladimir.shelomovskii@gmail.com, vvsss

## Video Solution

https://youtu.be/Mk-MCeVjSGc ~Shreyas S