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Difference between revisions of "2017 AIME I Problems/Problem 4"
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This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>.
 
This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>.
  
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==Solution 2==
+
==Shortcut==
−
From above, we find that the foot of the altitude from <math>P</math> lands on the circumcenter of <math>\triangle ABC</math>.  
+
Here is a shortcut for finding the radius <math>R</math> of the circumcenter of <math>\triangle ABC</math>.
 +
 
 +
As before, we find that the foot of the altitude from <math>P</math> lands on the circumcenter of <math>\triangle ABC</math>. Let <math>BC=a</math>, <math>AC=b</math>, and <math>AB=c</math>.
 
Then we write the area of <math>\triangle ABC</math> in two ways:
 
Then we write the area of <math>\triangle ABC</math> in two ways:
−
<cmath>[ABC] = </cmath>
+
<cmath>[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}</cmath>
 +
 
 +
Plugging in the values of <math>20</math>, <math>20</math>, and <math>24</math> for <math>a</math>, <math>b</math>, and <math>c</math> respectively, and solving for <math>R</math>, we obtain <math>R= \frac{25}{2}=OA=OB=OC</math>.
 +
 
 +
Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2017|n=I|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:58, 5 January 2018

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 the values of $20$, $20$, 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.

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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