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

2017 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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