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

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==Solution==
 
==Solution==
Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. We find that the altitude to side <math>AB</math> is <math>16</math>, so the area of <math>\triangle ABC</math> is <math>(24*16)/2 = 192</math>.
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Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. We find that the altitude to side <math>\overline {AB}</math> is <math>16</math>, so the area of <math>\triangle ABC</math> is <math>(24*16)/2 = 192</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>. Then,
 
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>. Then,
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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):
  
<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>
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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>.
  
Solution by Zeroman
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 +
NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows :
 +
 
 +
Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that <math>\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}</math>. Now integrate it taking the limits <math>0</math> to <math>H</math>
 +
 
 +
==Shortcut==
 +
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:
 +
<cmath>[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}</cmath>
 +
 
 +
Plugging in <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.
 +
 
 +
==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 <math>24</math> is at the origin, or <math>(0, 0, 0)</math>. Then, the two other vertices can be <math>(-12, -16, 0)</math> and <math>(12, -16, 0)</math>. Let the fourth vertex have coordinates of <math>(x, y, z)</math>. We have the following <math>3</math> equations from the distance formula.
 +
 
 +
<cmath>x^2+y^2+z^2=625</cmath>
 +
 
 +
<cmath>(x+12)^2+(y+16)^2+z^2=625</cmath>
 +
 
 +
<cmath>(x-12)^2+(y+16)^2+z^2=625</cmath>
 +
 
 +
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 <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'''
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 +
==Solution 3 (Heron's Formula)==
 +
 
 +
Label the four vertices of the tetrahedron and the midpoint of <math>\overline {AB}</math>, and notice that the area of the base of the tetrahedron, <math>\triangle ABC</math>, equals <math>192</math>, according to Solution 1.
 +
 
 +
Notice that the altitude of <math>\triangle CPM</math> from <math>\overline {CM}</math> to point <math>P</math> is the height of the tetrahedron. Side <math>\overline {PM}</math> is can be found using the Pythagorean Theorem on <math>\triangle APM</math>, giving us <math>\overline {PM}=\sqrt{481}.</math>
 +
 
 +
Using Heron's Formula, the area of <math>\triangle CPM</math> can be written as
 +
<cmath>\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})}</cmath>
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<cmath>=\frac{\sqrt{(41+\sqrt{481})(9+\sqrt{481})(-9+\sqrt{481})(41-\sqrt{481})}}{4}</cmath>
 +
 
 +
Notice that both <math>(41+\sqrt{481})(41-\sqrt{481})</math> and <math>(9+\sqrt{481})(-9+\sqrt{481})</math> can be rewritten as differences of squares; thus, the expression can be written as
 +
<cmath>\frac{\sqrt{(41^2-481)(481-9^2)}}{4}=\frac{\sqrt{480000}}{4}=100\sqrt{3}.</cmath>
 +
 
 +
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'''
 +
 
 +
==Video Solution==
 +
https://youtu.be/Mk-MCeVjSGc
 +
~Shreyas S
  
 
==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}}

Latest revision as of 05:59, 28 December 2020

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


NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows :

Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that $\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}$. Now integrate it taking the limits $0$ to $H$

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

Video Solution

https://youtu.be/Mk-MCeVjSGc ~Shreyas S

See Also

2017 AIME I (ProblemsAnswer KeyResources)
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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