2001 AIME I Problems/Problem 12
Contents
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 1
![[asy] import three; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label("$I$",I,W); label("$A$",A,S); label("$B$",B,S); label("$C$",C,W*-1); label("$D$",D,W*-1); [/asy]](http://latex.artofproblemsolving.com/a/d/5/ad592fa5d796f97d2c229a092105e33dc3dfc1ed.png)
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 tetrahedrons 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 ![$\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]$](http://latex.artofproblemsolving.com/b/a/0/ba0528715eb4dac76614710e6a3b9bb5a8ccc64a.png)
. We know triangle ABC's side lengths, 
and 
, so using the expanded form of heron's formula, ![\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*}](http://latex.artofproblemsolving.com/e/1/a/e1a74e0d4948d08dd4006b0d828d68e44a699e80.png)
Therefore, the surface area is 
, and the volume is ![$\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8$](http://latex.artofproblemsolving.com/3/8/b/38ba0074d10f2be358b0ddc91d100bce3cb8e212.png)
, 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 
.
Solution 5
Clearly, if the radius of the sphere is , the center of the sphere lies on 
.
We find the equation of plane to be 
. From the definition of the insphere, it must be true that the distance from the center of the sphere to plane is equal to the length of the radius of the sphere. By point-to-plane, we have ![\[r=\frac{|\frac16 r+\frac14 r+\frac12 r-1|}{\sqrt{\left(\frac16\right)^2+\left(\frac14\right)^2+\left(\frac12\right)^2}} \implies r=\frac23,\]](http://latex.artofproblemsolving.com/2/0/3/2038c9329b71447ed79523b0f6fc246cdcbfe195.png)
so the answer is .
-pqr.
Solution 6(Formula Bash)
The radius of the insphere in a tetrahedron can be calculated using the formula:
where \(r\) is the radius, \(V\) is the volume of the tetrahedron, and \(A\) is the total surface area.
The volume of the tetrahedron is calculated as:
The total surface area is the sum of the areas of the four triangular faces:
![$Surface Area = [ABD] + [BDC] + [ACD] + [ABC] = 12 + 4 + 6 + [ABC]$](http://latex.artofproblemsolving.com/5/f/7/5f7c91f0f5d0019f44c7bd70ef797401a460f7ad.png)
The side lengths of triangle \(ABC\) are:
\[
\sqrt{52}, \, \sqrt{40}, \, \sqrt{20}
\]
We construct an altitude from \(A\) and use the system of equations:
\[
x^2 + h^2 = 40
\]
\[
(2\sqrt{5} - x)^2 + h^2 = 20
\]
Solving this system gives:
\[
h = \frac{14\sqrt{5}}{5}
\]
The area of triangle \(ABC\) is:
\[
\text{Area of } [ABC] = 14
\]
Thus, the total surface area is:
\[
A = 12 + 4 + 6 + 14 = 36
\]
Finally, the radius of the insphere is:
\[
r = \frac{3 \times 8}{36} = \frac{2}{3}
\]
\[
\boxed{\frac{2}{3}}
\]
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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