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

Revision as of 13:10, 6 July 2019

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

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

Revision as of 13:10, 6 July 2019 (view source) Dzhou100 (talk \| contribs) ← Older edit		Revision as of 13:10, 6 July 2019 (view source) Dzhou100 (talk \| contribs) (→‎Solution 3 (Heron's Formula)) Newer edit →
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	From this, we can determine the height of both <math>\triangle CPM</math> and tetrahedron <math>ABCP</math> to be <math>\frac{100\sqrt{3}}{8}</math>; therefore, the volume of the tetrahedron equals <math>\frac{100\sqrt{3}}{8} \cdot 192=800\sqrt{3}</math>; thus, <math>m+n=800+3=\boxed{803}.</math>		From this, we can determine the height of both <math>\triangle CPM</math> and tetrahedron <math>ABCP</math> to be <math>\frac{100\sqrt{3}}{8}</math>; therefore, the volume of the tetrahedron equals <math>\frac{100\sqrt{3}}{8} \cdot 192=800\sqrt{3}</math>; thus, <math>m+n=800+3=\boxed{803}.</math>
		+
	-dzhou100		-dzhou100

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