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>. | + | 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 | + | 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, |
<cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath> | <cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath> | ||
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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 + ( | + | <cmath>25^2 = h^2 + (25/2)^2</cmath> |
<cmath>625 = h^2 + 625/4</cmath> | <cmath>625 = h^2 + 625/4</cmath> | ||
Line 47: | Line 47: | ||
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 | + | |
+ | 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''' | ||
+ | |||
+ | ==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''' | ||
+ | |||
+ | ==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
Contents
Problem 4
A pyramid has a triangular base with side lengths , , and . The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length . The volume of the pyramid is , where and are positive integers, and is not divisible by the square of any prime. Find .
Solution
Let the triangular base be , with . We find that the altitude to side is , so the area of is .
Let the fourth vertex of the tetrahedron be , and let the midpoint of be . Since is equidistant from , , and , the line through perpendicular to the plane of will pass through the circumcenter of , which we will call . Note that is equidistant from each of , , and . Then,
Let . Equation :
Squaring both sides, we have
Substituting with equation :
We now find that .
Let the distance . Using the Pythagorean Theorem on triangle , , or (all three are congruent by SSS):
Finally, by the formula for volume of a pyramid,
This simplifies to , so .
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 . Now integrate it taking the limits to
Shortcut
Here is a shortcut for finding the radius of the circumcenter of .
As before, we find that the foot of the altitude from lands on the circumcenter of . Let , , and . Then we write the area of in two ways:
Plugging in , , and for , , and respectively, and solving for , we obtain .
Then continue as before to use the Pythagorean Theorem on , find , 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 is at the origin, or . Then, the two other vertices can be and . Let the fourth vertex have coordinates of . We have the following equations from the distance formula.
Adding the last two equations and substituting in the first equation, we get that . If you drew a good diagram, it should be obvious that . Now, solving for , we get that . So, the height of the pyramid is . The base is equal to the area of the triangle, which is . The volume is . Thus, the answer is .
-RootThreeOverTwo
Solution 3 (Heron's Formula)
Label the four vertices of the tetrahedron and the midpoint of , and notice that the area of the base of the tetrahedron, , equals , according to Solution 1.
Notice that the altitude of from to point is the height of the tetrahedron. Side is can be found using the Pythagorean Theorem on , giving us
Using Heron's Formula, the area of can be written as
Notice that both and can be rewritten as differences of squares; thus, the expression can be written as
From this, we can determine the height of both and tetrahedron to be ; therefore, the volume of the tetrahedron equals ; thus,
-dzhou100
Video Solution
https://youtu.be/Mk-MCeVjSGc ~Shreyas S
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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