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

Latest revision as of 14:55, 5 January 2024

Problem

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 $\overline {OM} = d$ . Then $OC=OA=\sqrt{d^2+12^2}.$ 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.

Another Shortcut (Extended Law of Sines)

Take the base $\triangle ABC$ , where $AB = BC = 20$ and $AC = 24$ . Draw an altitude from $B$ to $AC$ that bisects $AC$ at point $D$ . Then the altitude has length $\sqrt{20^2 - 12^2} = \sqrt{16^2} = 16$ . Next, let $\angle BCA = \theta$ . Then from the right triangle $\triangle BDC$ , $\sin \theta = 4/5$ . From the extended law of sines, the circumradius is $20 \cdot \dfrac{5}{4} \cdot \dfrac{1}{2} = \dfrac{25}{2}$ .

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

Solution 4 (Symmetry)

Notation is shown on diagram. $AM = MB = c = 12, AC = BC = b = 20,$ $DA = DB = DC = a = 25.$ $CM = x + y = \sqrt{b^2-c^2} = 16,$ $x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,$ $x – y = \frac{x^2 – y^2}{x+y} = \frac {c^2} {16} = 9,$ $x = \frac {16 + 9}{2} = \frac {a}{2},$ $h = \sqrt{a^2 -\frac{ a^2}{4}} = a \frac {\sqrt{3}}{2},$ $V = \frac{h\cdot CM \cdot c}{3}= \frac{16\cdot 25 \sqrt{3} \cdot 12}{3} = 800 \sqrt{3} \implies \boxed {803}.$ vladimir.shelomovskii@gmail.com, vvsss

Video Solution

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

@@ Line 1: / Line 1: @@
-==Problem 4==
+==Problem ==
 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==
-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>.
+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>. We find that <math>\overline {CM} = 16</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,
-<math>\overline {OM} + \overline {OC} = \overline {CM} = 16</math>
+<cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath>
-<math>d + \sqrt {d^2 + 144} = 16</math> (1)
+Let <math>\overline {OM} = d</math>. Then <math>OC=OA=\sqrt{d^2+12^2}.</math>
+Equation <math>(1)</math>:
+<cmath>d + \sqrt {d^2 + 144} = 16</cmath>
 Squaring both sides, we have
-<math>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</math>
+<cmath>d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256</cmath>
-<math>2d^2 + 2d\sqrt {d^2+144} = 112</math>
+<cmath>2d^2 + 2d\sqrt {d^2+144} = 112</cmath>
-<math>2d(d + \sqrt {d^2+144}) = 112</math>
+<cmath>2d(d + \sqrt {d^2+144}) = 112</cmath>
-Substituting with equation (1):
+Substituting with equation <math>(1)</math>:
-<math>2d(16) = 112</math>
+<cmath>2d(16) = 112</cmath>
-<math>d = 7/2</math>.
+<cmath>d = 7/2</cmath>
 We now find that <math>\sqrt{d^2 + 144} = 25/2</math>.
@@ Line 28: / Line 31: @@
 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>
+<cmath>25^2 = h^2 + (25/2)^2</cmath>
-<math>625 = h^2 + 625/4</math>
+<cmath>625 = h^2 + 625/4</cmath>
-<math>1875/4 = h^2</math>
+<cmath>1875/4 = h^2</cmath>
-<math>25\sqrt {3} / 2 = h</math>.
+<cmath>25\sqrt {3} / 2 = h</cmath>
 Finally, by the formula for volume of a pyramid,
-<math>V = Bh/3</math>
+<cmath>V = Bh/3</cmath>
+<cmath>V = (192)(25\sqrt{3}/2)/3</cmath>
+This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>.
+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.
+===Another Shortcut (Extended Law of Sines)===
+Take the base <math>\triangle ABC</math>, where <math>AB = BC = 20</math> and <math>AC = 24</math>. Draw an altitude from <math>B</math> to <math>AC</math> that bisects <math>AC</math> at point <math>D</math>. Then the altitude has length <math>\sqrt{20^2 - 12^2} = \sqrt{16^2} = 16</math>. Next, let <math>\angle BCA = \theta</math>. Then from the right triangle <math>\triangle BDC</math>, <math>\sin \theta = 4/5</math>. From the extended law of sines, the circumradius is <math>20 \cdot \dfrac{5}{4} \cdot \dfrac{1}{2} = \dfrac{25}{2}</math>.
+==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'''
+==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>
+<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'''
+==Solution 4 (Symmetry)==
+[[File:2017 AIME I 4.png|450px|right]]
+Notation is shown on diagram.
+<cmath>AM = MB = c = 12, AC = BC = b = 20,</cmath>
+<cmath>DA = DB = DC = a = 25.</cmath>
+<cmath>CM = x + y = \sqrt{b^2-c^2} = 16,</cmath>
+<cmath>x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,</cmath>
+<cmath>x – y = \frac{x^2 – y^2}{x+y} = \frac {c^2} {16} = 9,</cmath>
+<cmath>x = \frac {16 + 9}{2} = \frac {a}{2},</cmath>
+<cmath>h = \sqrt{a^2 -\frac{ a^2}{4}} = a \frac {\sqrt{3}}{2},</cmath>
+<cmath>V = \frac{h\cdot CM \cdot c}{3}= \frac{16\cdot 25 \sqrt{3} \cdot 12}{3} = 800 \sqrt{3} \implies \boxed {803}.</cmath>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Video Solution==
+https://youtu.be/Mk-MCeVjSGc
+~Shreyas S
-<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>.
+==See Also==
+{{AIME box|year=2017|n=I|num-b=3|num-a=5}}
+{{MAA Notice}}

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