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

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A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
 
A pyramid has a triangular base with side lengths <math>20</math>, <math>20</math>, and <math>24</math>. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length <math>25</math>. The volume of the pyramid is <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
 
==Solution==
 
==Solution==
−
Let the triangular base be <math>\triangle ABC</math>. Using Simplified Heron's formula for the area of an isosceles triangle gives <math>12\sqrt{32(8)}=192</math>.
+
Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. Using Simplified Heron's formula for the area of an isosceles triangle gives <math>12\sqrt{32(8)}=192</math>.
  
−
<math>\vdots</math>
+
Let the fourth vertex of the tetrahedron be <math>P</math>, and let the midpoint of <math>\overline {AB}</math> be <math>M</math>. Since <math>P</math> is equidistant from <math>A</math>, <math>B</math>, and <math>C</math>, the line through <math>P</math> perpendicular to the plane of <math>\triangle ABC</math> will pass through the circumcenter of <math>\triangle ABC</math>, which we will call <math>O</math>. Note that <math>O</math> is equidistant from each of <math>A</math>, <math>B</math>, and <math>C</math>. We find that <math>\overline {CM} = 16</math>. Then,
  
−
The volume of the pyramid is <math>800\sqrt{3}</math>, so the answer is <math>\boxed{803}</math>
+
<math>\overline {OM} + \overline {OC} = \overline {CM} = 16</math>
 +
 
 +
<math>d + \sqrt {d^2 + 144} = 16</math> (1)
 +
 
 +
Squaring both sides, we have
 +
 
 +
<math>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</math>
 +
 
 +
<math>2d^2 + 2d\sqrt {d^2+144} = 112</math>
 +
 
 +
<math>2d(d + \sqrt {d^2+144}) = 112</math>
 +
 
 +
Substituting with equation (1):
 +
 
 +
<math>2d(16) = 112</math>
 +
 
 +
<math>d = 7/2</math>.
 +
 
 +
We now find that <math>\sqrt{d^2 + 144} = 25/2</math>.
 +
 
 +
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):
 +
 
 +
<math>25^2 = h^2 + (\sqrt {d^2 + 144})^2</math>
 +
 
 +
<math>625 = h^2 + 625/4</math>
 +
 
 +
<math>1875/4 = h^2</math>
 +
 
 +
<math>25\sqrt {3} / 2 = h</math>.
 +
 
 +
 
 +
Finally, by the formula for volume of a pyramid,
 +
 
 +
<math>V = Bh/3</math>
 +
 
 +
<math>V = (192)(25\sqrt{3}/2)/3</math>. This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>.

Revision as of 16:40, 8 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 $\overline {AB} = 24$. Using Simplified Heron's formula for the area of an isosceles triangle gives $12\sqrt{32(8)}=192$.

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 $C$. We find that $\overline {CM} = 16$. Then,

$\overline {OM} + \overline {OC} = \overline {CM} = 16$

$d + \sqrt {d^2 + 144} = 16$ (1)

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):

$25^2 = h^2 + (\sqrt {d^2 + 144})^2$

$625 = h^2 + 625/4$

$1875/4 = h^2$

$25\sqrt {3} / 2 = h$.


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

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