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

Revision as of 17:44, 27 September 2023

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 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$",D,W*-1); [/asy]$

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 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 $V$ 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, $\begin{align*}[ABC]&=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2})-a^{4}-b^{4}-c^{4}}{16}}\\ &=\sqrt{2(5\cdot{13}+10\cdot{5}+13\cdot{10})-5^{2}-10^{2}-13^{2}}\\ &=\sqrt{196}\\ &=14\end{align*}$ 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

Solution 4

First let us find the equation of the plane passing through $(6,0,0), (0,0,2), (0,4,0)$ . The "point-slope form" is $A(6-x1)+B(0-y1)+C(0-z1)=0.$ Plugging in $(0,0,2)$ gives $A(6)+B(0)+C(-2)=0.$ Plugging in $(0,4,0)$ gives $A(6)+B(-4)+C(0)=0.$ We can then use Cramer's rule/cross multiplication to get $A/(0-8)=-B/(0+12)=C/(-24)=k.$ Solve for A, B, C to get $2k, 3k, 6k$ respectively. We can then get $2k(x-x1)+3k(y-y1)+6k(z-z1)=0.$ Cancel out k on both sides. Next, let us substitute $(0,0,2)$ . We can then get $2x+3y+6z=12$ as the equation of the plane. We can divide the equation by its magnitude to get the normal form of the plane. We get $2x/7+2y/7+6z/7=12/7$ to be the normal form. Note that the point is going to be at $(r,r,r).$ We find the distance from $(r,r,r)$ to the plane as $2/7r+3/7r+6/7r-12/7/(\sqrt{(4/49+9/49+36/49)})$ , which is $+/-(11r/7-12/7)$ . We take the negative value of this because if we plug in $(0,0,0)$ to the equation of the plane we get a negative value. We equate that value to r and we get the equation $-(11r/7-12/7)=r$ to solve $r={2/3}$ , so the answer is $\boxed{005}$ .

Solution 5

Clearly, if the radius of the sphere is $r$ , the center of the sphere lies on $(r, r, r)$ .

We find the equation of plane $ABC$ to be $\frac16 x+\frac14 y+\frac12 z=1$ . From the definition of the insphere, it must be true that the distance from the center of the sphere to plane $ABC$ is equal to the length of the radius of the sphere. By point-to-plane, we have $r=\frac{|\frac16 r+\frac14 r+\frac12 r-1|}{\sqrt{\left(\frac16\right)^2+\left(\frac14\right)^2+\left(\frac12\right)^2}} \implies r=\frac23,$ so the answer is $\boxed{005}$ .

-pqr.

@@ Line 2: / Line 2: @@
 A [[sphere]] is inscribed in the [[tetrahedron]] whose vertices are <math>A = (6,0,0), B = (0,4,0), C = (0,0,2),</math> and <math>D = (0,0,0).</math>  The [[radius]] of the sphere is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m + n.</math>
-== Solution ==
+== Solution 1==
-<center><asy>
+<asy>
-import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(5,-10,4);
+import three;
-pair A=(6,0,0), B=(0,4,0), C=(0,0,2), D=(0,0,0);
+currentprojection = perspective(-2,9,4);
-D(MP("A",A)--MP("B",B)--MP("C",C,N)--A--MP("D",D)--B--D--C);
+triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0);
-</asy></center>
+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>
-Connect all four vertices of tetrahedron <math>ABCD</math> to its [[incenter]], <math>I</math>. This yields four tetrahedra <math>ABCI, ABDI, ACDI, BCDI</math>, all of which have height of <math>r</math> (the radius of the inscribed sphere), and which together form <math>ABCD</math>. It follows that <center><math>\begin{align*}V &= V_{ABCI}+V_{ABDI}+V_{ACDI}+V_{BCDI}\\ &= \frac 13 \cdot r \cdot \left([ABC] + [ABD] + [ACD] + [BCD]\right) = \frac{rS}{3}, \end{align*}</math></center> where <math>S</math> is the [[surface area]] of <math>ABCD</math>.
+The center <math>I</math> of the insphere must be located at <math>(r,r,r)</math> where <math>r</math> is the sphere's radius.
+<math>I</math> must also be a distance <math>r</math> from the plane <math>ABC</math>
-Since <math>\triangle ABD, ACD, BCD</math> all lie on the planes containing the axes, their areas are straightforward to calculate; respectively <math>12,6,4</math>. To find <math>[ABC]</math>, we can use the 3-dimensional [[distance formula]] (<math>d=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}</math>) to find that <math>AB = \sqrt{52}, BC=\sqrt{20}, CA=\sqrt{40}</math>. From here, we can use the [[Law of Cosines]] and the sine area formula to compute <math>[ABC]</math>, or we can use a manipulated version of [[Heron's formula]]: <math>A = \frac{1}{4}\sqrt{4a^2b^2 - (a^2+b^2-c^2)^2} = 14</math>.{{ref|1}}
+The signed distance between a plane and a point <math>I</math> can be calculated as <math>\frac{(I-G) \cdot P}{|P|}</math>, where G is any point on the plane, and P is a vector perpendicular to ABC.
-Thus, <math>S = 14 + 12 + 6 + 4 = 36</math>. The volume of <math>ABCD</math> we can compute by letting <math>AD</math> to be the height to face <math>BCD</math>, so <math>V = \frac{1}{3} \cdot 6 \cdot \frac 12 \cdot 4 \cdot 2 = 8</math>. Therefore, <math>r = \frac{3V}{S} = \frac{24}{36} = \frac 23</math>, and <math>m+n = \boxed{005}</math>.
+A vector <math>P</math> perpendicular to plane <math>ABC</math> can be found as <math>V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle</math>
+Thus <math>\frac{(I-C) \cdot P}{|P|}=-r</math> where the negative comes from the fact that we want <math>I</math> to be in the opposite direction of <math>P</math>
+<cmath>\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\\
+r-48&=-28r\\
+r&=48\\
+r&=\frac{2}{3}
+\end{align*}</cmath>
-<div style="font-size:80%">{{note|1}} There are a couple of other ways to compute <math>[ABC]</math>, including by [[vector]]s. In fact, it is known that in a trirectangular tetrahedron (one in which three edges are mutually perpendicular, as is the case here), the sum of the squares of the areas of the three smaller faces equals the square of the area of the larger face. See the thread below for details. </div>
+Finally <math>2+3=\boxed{005}</math>
+==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 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 <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 3==
+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.
+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.
+-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>.
+==Solution 5==
+Clearly, if the radius of the sphere is <math>r</math>, the center of the sphere lies on <math>(r, r, r)</math>.
+We find the equation of plane <math>ABC</math> to be <math>\frac16 x+\frac14 y+\frac12 z=1</math>. From the definition of the insphere, it must be true that the distance from the center of the sphere to plane <math>ABC</math> is equal to the length of the radius of the sphere. By point-to-plane, we have <cmath>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,</cmath> so the answer is <math>\boxed{005}</math>.
+-pqr.
 == See also ==
-*<url>viewtopic.php?p=384205#384205 Discussion on AoPS</url>
 {{AIME box|year=2001|n=I|num-b=11|num-a=13}}
 [[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}

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