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

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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 therefore $\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}$ . Thus, the answer is $800+3 = \boxed{803}$ .

-RootThreeOverTwo

@@ Line 70: / Line 70: @@
 Adding the last two equations and substituting in the first equation, we get that <math>y=-\frac{25}{2}</math>. If you drew a good diagram, it should be obvious that <math>x=0</math>. Now, solving for <math>z</math>, we get that <math>z=\frac{25\sqrt{3}}{2}</math>. So, the height of the pyramid is <math>\frac{25\sqrt{3}}{2}</math>. The base is equal to the area of the triangle, which is <math>\frac{1}{2} \cdot 24 \cdot 16 = 192</math>. The volume is therefore <math>\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}</math>. Thus, the answer is <math>800+3 = \boxed{803}</math>.
--RootThreeOverTwo
+'''-RootThreeOverTwo'''
 ==See Also==
 {{AIME box|year=2017|n=I|num-b=3|num-a=5}}
 {{MAA Notice}}

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