Difference between revisions of "2001 AIME I Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
− | Notice that we can split the tetrahedron into <math>4</math> smaller tetrahedrons such that the height of each tetrahedron is <math>r</math> and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be <math>V</math> and surface area be <math>F</math>, using the volume formula for each pyramid(base times height divided by 3) we have <math>\dfrac{rF}{3}=V</math>. The surface area of the pyramid is <math>\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]</math>. We know triangle ABC's side lengths, <math>\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},</math> and <math>\sqrt{4^{2}+6^{2}}</math>, so using the expanded form of heron's formula, < | + | Notice that we can split the tetrahedron into <math>4</math> smaller tetrahedrons such that the height of each tetrahedron is <math>r</math> and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be <math>V</math> and surface area be <math>F</math>, using the volume formula for each pyramid(base times height divided by 3) we have <math>\dfrac{rF}{3}=V</math>. The surface area of the pyramid is <math>\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]</math>. We know triangle ABC's side lengths, <math>\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},</math> and <math>\sqrt{4^{2}+6^{2}}</math>, so using the expanded form of heron's formula, <cmath>\begin{align*}[ABC]&=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}\\ |
+ | &=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}\\ | ||
+ | &=\sqrt{196}\\ | ||
+ | &=14\end{align*}</cmath> | ||
+ | Therefore, the surface area is <math>14+22=36</math>, and the volume is <math>\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8</math>, and using the formula above that <math>\dfrac{rF}{3}=V</math>, we have <math>12r=8</math> and thus <math>r=\dfrac{2}{3}</math>, so the desired answer is <math>2+3=\boxed{005}</math>. | ||
(Solution by Shaddoll) | (Solution by Shaddoll) | ||
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-JZ | -JZ | ||
+ | |||
+ | ==Solution 4== | ||
+ | First let us find the equation of the plane passing through <math>(6,0,0), (0,0,2), (0,4,0)</math>. The "point-slope form" is <math>A(6-x1)+B(0-y1)+C(0-z1)=0.</math> Plugging in <math>(0,0,2)</math> gives <math>A(6)+B(0)+C(-2)=0.</math> Plugging in <math>(0,4,0)</math> gives <math>A(6)+B(-4)+C(0)=0.</math> We can then use Cramer's rule/cross multiplication to get <math>A/(0-8)=-B/(0+12)=C/(-24)=k.</math> Solve for A, B, C to get <math>2k, 3k, 6k</math> respectively. We can then get <math>2k(x-x1)+3k(y-y1)+6k(z-z1)=0.</math> Cancel out k on both sides. Next, let us substitute <math>(0,0,2)</math>. We can then get <math>2x+3y+6z=12 </math>as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get <math>2x/7+2y/7+6z/7=12/7</math> to be the normal form. Note that the point is going to be at <math>(r,r,r).</math> We find the distance from <math>(r,r,r)</math> to the plane as <math>2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})</math>, which is <math>+/-(11r/7-12/7)</math>. We take the negative value of this because if we plug in <math>(0,0,0)</math> to the equation of the plane we get a negative value. We equate that value to r and we get the equation <math>-(11r/7-12/7)=r</math> to solve <math>r={2/3}</math>, so the answer is <math>\boxed{005}</math>. | ||
== See also == | == See also == | ||
− | |||
{{AIME box|year=2001|n=I|num-b=11|num-a=13}} | {{AIME box|year=2001|n=I|num-b=11|num-a=13}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:34, 26 December 2021
Problem
A sphere is inscribed in the tetrahedron whose vertices are and The radius of the sphere is where and are relatively prime positive integers. Find
Solution
The center of the insphere must be located at where is the sphere's radius. must also be a distance from the plane
The signed distance between a plane and a point can be calculated as , where G is any point on the plane, and P is a vector perpendicular to ABC.
A vector perpendicular to plane can be found as
Thus where the negative comes from the fact that we want to be in the opposite direction of
Finally
Solution 2
Notice that we can split the tetrahedron into smaller tetrahedrons such that the height of each tetrahedron is and the base of each tetrahedron is one of the faces of the original tetrahedron. This is because the bases of the spheres are tangent to the sphere, so the line from the center to the foot of the perpendicular to the bases hits the tangency points. Letting volume be and surface area be , using the volume formula for each pyramid(base times height divided by 3) we have . The surface area of the pyramid is . We know triangle ABC's side lengths, and , so using the expanded form of heron's formula, Therefore, the surface area is , and the volume is , and using the formula above that , we have and thus , so the desired answer is .
(Solution by Shaddoll)
Solution 3
The intercept form equation of the plane is Its normal form is (square sum of the coefficients equals 1). The distance from to the plane is . Since and are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in ). Therefore we have So which solves the problem.
Additionally, if is on the other side of , we have , which yields corresponding an "ex-sphere" that is tangent to face as well as the extensions of the other 3 faces.
-JZ
Solution 4
First let us find the equation of the plane passing through . The "point-slope form" is Plugging in gives Plugging in gives We can then use Cramer's rule/cross multiplication to get Solve for A, B, C to get respectively. We can then get Cancel out k on both sides. Next, let us substitute . We can then get as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get to be the normal form. Note that the point is going to be at We find the distance from to the plane as , which is . We take the negative value of this because if we plug in to the equation of the plane we get a negative value. We equate that value to r and we get the equation to solve , so the answer is .
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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