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Difference between revisions of "2001 AIME I Problems/Problem 12"

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(Solution by Shaddoll)
 
(Solution by Shaddoll)
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==Solution 3==
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The intercept form equation of the plane <math>ABC</math> is <math>\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.</math> Its normal form is <math>\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0</math> (square sum of the coefficients equals 1). The distance from <math>(r,r,r)</math> to the plane is <math>\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |</math>. Since <math>(r,r,r)</math> and <math>(0,0,0)</math> are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in <math>(0,0,0)</math>). Therefore we have
 +
<math>-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.</math> So <math>r=\dfrac{2}{3},</math> which solves the problem.
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 +
Additionally, if <math>(r,r,r)</math> is on the other side of <math>ABC</math>, we have <math>\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r</math>, which yields <math>r=\dfrac{12}{5},</math> corresponding an "ex-sphere" that is tangent to face <math>ABC</math> as well as the extensions of the other 3 faces.
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-JZ
  
 
== See also ==
 
== See also ==

Revision as of 13:17, 11 February 2020

Problem

A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

[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]

The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$

The signed distance between a plane and a point $I$ can be calculated as $\frac{(I-G) \cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to ABC.

A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle$

Thus $\frac{(I-C) \cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$

\begin{align*}\frac{(I-C) \cdot P}{|P|}&=-r\\ \frac{(\langle r, r, r \rangle-\langle 0, 0, 2 \rangle) \cdot P}{|P|}&=-r\\ \frac{\langle r, r, r-2 \rangle \cdot \langle 8, 12, 24 \rangle}{\langle 8, 12, 24 \rangle}&=-r\\ \frac{44r -48}{28}&=-r\\ 44r-48&=-28r\\ 72r&=48\\ r&=\frac{2}{3} \end{align*}


Finally $2+3=\boxed{005}$

Solution 2

Notice that we can split the tetrahedron into $4$ smaller tetrahedrons such that the height of each tetrahedron is $r$ 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 $V$ and surface area be $F$, using the volume formula for each pyramid(base times height divided by 3) we have $\dfrac{rF}{3}=V$. The surface area of the pyramid is $\dfrac{6\cdot{4}+6\cdot{2}+4\cdot{2}}{2}+[ABC]=22+[ABC]$. We know triangle ABC's side lengths, $\sqrt{2^{2}+4^{2}}, \sqrt{2^{2}+6^{2}},$ and $\sqrt{4^{2}+6^{2}}$, so using the expanded form of heron's formula, $[ABC]=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}=\sqrt{196}=14$. Therefore, the surface area is $14+22=36$, and the volume is $\dfrac{[BCD]\cdot{6}}{3}=\dfrac{4\cdot{2}\cdot{6}}{3\cdot{2}}=8$, and using the formula above that $\dfrac{rF}{3}=V$, we have $12r=8$ and thus $r=\dfrac{2}{3}$, so the desired answer is $2+3=\boxed{005}$.

(Solution by Shaddoll)

Solution 3

The intercept form equation of the plane $ABC$ is $\frac{x}{6}+\dfrac{y}{4}+\dfrac{z}{2}=1.$ Its normal form is $\dfrac{2}{7}x+\dfrac{3}{7}y+\dfrac{6}{7}z-\dfrac{12}{7}=0$ (square sum of the coefficients equals 1). The distance from $(r,r,r)$ to the plane is $\left |\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right |$. Since $(r,r,r)$ and $(0,0,0)$ are on the same side of plane, the value in the absolute value sign is negative (same as the one by plugging in $(0,0,0)$). Therefore we have $-\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r.$ So $r=\dfrac{2}{3},$ which solves the problem.

Additionally, if $(r,r,r)$ is on the other side of $ABC$, we have $\left (\dfrac{2}{7}r+\dfrac{3}{7}r+\dfrac{6}{7}r-\dfrac{12}{7}\right )=r$, which yields $r=\dfrac{12}{5},$ corresponding an "ex-sphere" that is tangent to face $ABC$ as well as the extensions of the other 3 faces.

-JZ

See also

  • <url>viewtopic.php?p=384205#384205 Discussion on AoPS</url>
2001 AIME I (Problems • Answer Key • Resources)
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