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

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Let the distance <math>\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by SSS):
 
Let the distance <math>\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by SSS):
  
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<cmath>25^2 = h^2 + (\sqrt {d^2 + 144})^2</cmath>
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<cmath>25^2 = h^2 + (25/2)^2</cmath>
  
 
<cmath>625 = h^2 + 625/4</cmath>
 
<cmath>625 = h^2 + 625/4</cmath>

Revision as of 20:12, 20 March 2017

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

Solution by Zeroman

See Also

2017 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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