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+ | ==Problem== | ||
+ | Let <math>ABCD</math> be a tetrahedron such that <math>AB=CD= \sqrt{41}</math>, <math>AC=BD= \sqrt{80}</math>, and <math>BC=AD= \sqrt{89}</math>. There exists a point <math>I</math> inside the tetrahedron such that the distances from <math>I</math> to each of the faces of the tetrahedron are all equal. This distance can be written in the form <math>\frac{m \sqrt n}{p}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+p</math>. | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
− | Notice that | + | Notice that \(41=4^2+5^2\), \(89=5^2+8^2\), and \(80=8^2+4^2\), let \(A~(0,0,0)\), \(B~(4,5,0)\), \(C~(0,5,8)\), and \(D~(4,0,8)\). Then the plane \(BCD\) has a normal |
− | + | \begin{equation*} | |
\mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}. | \mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}. | ||
− | + | \end{equation*} | |
− | Hence, the distance from | + | Hence, the distance from \(A\) to plane \(BCD\), or the height of the tetrahedron, is |
− | + | \begin{equation*} | |
h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}. | h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}. | ||
− | + | \end{equation*} | |
− | Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it | + | Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it \(S\). Then by the volume formula for pyramids, |
\begin{align*} | \begin{align*} | ||
\frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\ | \frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\ | ||
&=\frac13Sr\cdot4. | &=\frac13Sr\cdot4. | ||
\end{align*} | \end{align*} | ||
− | Hence, | + | Hence, \(r=\tfrac h4=\tfrac{20\sqrt{21}}{63}\), and so the answer is \(20+21+63=\boxed{104}\). |
<font size=2>Solution by Quantum-Phantom</font> | <font size=2>Solution by Quantum-Phantom</font> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <asy> | ||
+ | import three; | ||
+ | |||
+ | currentprojection = orthographic(1,1,1); | ||
+ | |||
+ | triple O = (0,0,0); | ||
+ | triple A = (0,2,0); | ||
+ | triple B = (0,0,1); | ||
+ | triple C = (3,0,0); | ||
+ | triple D = (3,2,1); | ||
+ | triple E = (3,2,0); | ||
+ | triple F = (0,2,1); | ||
+ | triple G = (3,0,1); | ||
+ | |||
+ | draw(A--B--C--cycle, red); | ||
+ | draw(A--B--D--cycle, red); | ||
+ | draw(A--C--D--cycle, red); | ||
+ | draw(B--C--D--cycle, red); | ||
+ | |||
+ | draw(E--A--O--C--cycle); | ||
+ | draw(D--F--B--G--cycle); | ||
+ | draw(O--B); | ||
+ | draw(A--F); | ||
+ | draw(E--D); | ||
+ | draw(C--G); | ||
+ | |||
+ | label("$O$", O, SW); | ||
+ | label("$A$", A, NW); | ||
+ | label("$B$", B, W); | ||
+ | label("$C$", C, S); | ||
+ | label("$D$", D, NE); | ||
+ | label("$E$", E, SE); | ||
+ | label("$F$", F, NW); | ||
+ | label("$G$", G, NE); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | Inscribe tetrahedron <math>ABCD</math> in an rectangular prism as shown above. | ||
+ | |||
+ | By the Pythagorean theorem, we note | ||
+ | |||
+ | <cmath>OA^2 + OB^2 = AB^2 = 41,</cmath> | ||
+ | <cmath>OA^2 + OC^2 = AC^2 = 80, \text{and}</cmath> | ||
+ | <cmath>OB^2 + OC^2 = BC^2 = 89.</cmath> | ||
+ | |||
+ | Solving yields <math>OA = 4, OB = 5,</math> and <math>OC = 8.</math> | ||
+ | |||
+ | Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of <math>ABCD.</math> We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of <math>ABCD.</math> | ||
+ | |||
+ | We know that the distance from all <math>4</math> faces must be the same, so we only need to find the distance from the center to plane <math>ABC</math>. | ||
+ | |||
+ | Let <math>O = (0,0,0), A = (4,0,0), B = (0,5,0),</math> and <math>C = (0,0,8).</math> We obtain that the plane of <math>ABC</math> can be marked as <math>\frac{x}{4} + \frac{y}{5} + \frac{z}{8} = 1,</math> or <math>10x + 8y + 5z - 40 = 0,</math> and the center of the prism is <math>(2,\frac{5}{2},4).</math> | ||
+ | |||
+ | Using the Point-to-Plane distance formula, our distance is | ||
+ | |||
+ | <cmath>d = \frac{|10\cdot 2 + 8\cdot \frac{5}{2} + 5\cdot 4 - 40|}{\sqrt{10^2 + 8^2 + 5^2}} = \frac{20}{\sqrt{189}} = \frac{20\sqrt{21}}{63}.</cmath> | ||
+ | |||
+ | Our answer is <math>20 + 21 + 63 = \boxed{104}.</math> | ||
+ | |||
+ | - [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8] | ||
+ | |||
+ | ==Solution 3(Formula Abuse)== | ||
+ | We use the formula for the volume of iscoceles tetrahedron. | ||
+ | <math>V = \sqrt{(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)/72}</math> | ||
+ | |||
+ | Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find <cmath>\cos{\angle ACB} = \frac{80 + 89 - 41}{2\sqrt{80\cdot 89}}= \frac{16}{9\sqrt{5}}.</cmath>. | ||
+ | |||
+ | From this, we find <cmath>\sin{\angle ACB} = \sqrt{1-\cos^2{\angle ACB}} = \sqrt{1 - \frac{256}{405}} = \sqrt{\frac{149}{405}}</cmath> and can find the area of <math>\triangle ABC</math> as <cmath>A = \frac{1}{2} \sqrt{89\cdot 80}\cdot \sin{\angle ACB} = 6\sqrt{21}.</cmath> | ||
+ | |||
+ | Let <math>R</math> be the distance we want to find. By taking the sum of (equal) volumes <cmath>[ABCI] + [ABDI] + [ACDI] + [BCDI] = V,</cmath> We have <cmath>V = \frac{4AR}{3}.</cmath> Plugging in and simplifying, we get <math>R = \frac{20\sqrt{21}}{63}</math> for an answer of <math>20 + 21 + 63 = \boxed{104}</math> | ||
+ | |||
+ | ~AtharvNaphade | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>AH</math> be perpendicular to <math>BCD</math> that meets this plane at point <math>H</math>. | ||
+ | Let <math>AP</math>, <math>AQ</math>, and <math>AR</math> be heights to lines <math>BC</math>, <math>CD</math>, and <math>BD</math> with feet <math>P</math>, <math>Q</math>, and <math>R</math>, respectively. | ||
+ | |||
+ | We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as <math>A</math>, is <math>A = 6 \sqrt{21}</math>. | ||
+ | |||
+ | Hence, by using this area, we can compute <math>AP</math>, <math>AQ</math> and <math>AR</math>. | ||
+ | We have <math>AP = \frac{2 A}{BC} = \frac{2A}{\sqrt{89}}</math>, <math>AQ = \frac{2 A}{CD} = \frac{2A}{\sqrt{41}}</math>, and <math>AR = \frac{2 A}{BC} = \frac{2A}{\sqrt{80}}</math>. | ||
+ | |||
+ | Because <math>AH \perp BCD</math>, we have <math>AH \perp BC</math>. Recall that <math>AP \perp BC</math>. | ||
+ | Hence, <math>BC \perp APH</math>. Hence, <math>BC \perp HP</math>. | ||
+ | |||
+ | Analogously, <math>CD \perp HQ</math> and <math>BD \perp HR</math>. | ||
+ | |||
+ | We introduce a function <math>\epsilon \left( l \right)</math> for <math>\triangle BCD</math> that is equal to 1 (resp. -1) if point <math>H</math> and the opposite vertex of side <math>l</math> are on the same side (resp. opposite sides) of side <math>l</math>. | ||
+ | |||
+ | The area of <math>\triangle BCD</math> is | ||
+ | \begin{align*} | ||
+ | A & = \epsilon_{BC} {\rm Area} \ \triangle HBC | ||
+ | + \epsilon_{CD} {\rm Area} \ \triangle HCD | ||
+ | + \epsilon_{BD} {\rm Area} \ \triangle HBD \\ | ||
+ | & = \frac{1}{2} \epsilon_{BC} BC \cdot HP | ||
+ | + \frac{1}{2} \epsilon_{CD} CD \cdot HQ + | ||
+ | \frac{1}{2} \epsilon_{BD} BD \cdot HR \\ | ||
+ | & = \frac{1}{2} \epsilon_{BC} BC \cdot \sqrt{AP^2 - AH^2} | ||
+ | + \frac{1}{2} \epsilon_{CD} CD \cdot \sqrt{AQ^2 - AH^2} \\ | ||
+ | & \quad + \frac{1}{2} \epsilon_{BD} CD \cdot \sqrt{AR^2 - AH^2} . \hspace{1cm} (1) | ||
+ | \end{align*} | ||
+ | |||
+ | Denote <math>B = 2A</math>. | ||
+ | The above equation can be organized as | ||
+ | \begin{align*} | ||
+ | B & = \epsilon_{BC} \sqrt{B^2 - 89 AH^2} | ||
+ | + \epsilon_{CD} \sqrt{B^2 - 41 AH^2} \\ | ||
+ | & \quad + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} . | ||
+ | \end{align*} | ||
+ | |||
+ | This can be further reorganized as | ||
+ | \begin{align*} | ||
+ | B - \epsilon_{BC} \sqrt{B^2 - 89 AH^2} | ||
+ | & = \epsilon_{CD} \sqrt{B^2 - 41 AH^2} | ||
+ | + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} . | ||
+ | \end{align*} | ||
+ | |||
+ | Taking squares on both sides and reorganizing terms, we get | ||
+ | \begin{align*} | ||
+ | & 16 AH^2 - \epsilon_{BC} B \sqrt{B^2 - 89 AH^2} \\ | ||
+ | & = \epsilon_{CD} \epsilon_{BD} | ||
+ | \sqrt{\left( B^2 - 41 AH^2 \right) \left( B^2 - 80 AH^2 \right)} . | ||
+ | \end{align*} | ||
+ | |||
+ | Taking squares on both sides and reorganizing terms, we get | ||
+ | <cmath> | ||
+ | \[ | ||
+ | - \epsilon_{BC} 2 B \sqrt{B^2 - 89 AH^2} = - 2 B^2 + 189 AH^2 . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Taking squares on both sides, we finally get | ||
+ | \begin{align*} | ||
+ | AH & = \frac{20B}{189} \\ | ||
+ | & = \frac{40A}{189}. | ||
+ | \end{align*} | ||
+ | |||
+ | Now, we plug this solution to Equation (1). We can see that <math>\epsilon_{BC} = -1</math>, <math>\epsilon_{CD} = \epsilon_{BD} = 1</math>. | ||
+ | This indicates that <math>H</math> is out of <math>\triangle BCD</math>. To be specific, <math>H</math> and <math>D</math> are on opposite sides of <math>BC</math>, <math>H</math> and <math>C</math> are on the same side of <math>BD</math>, and <math>H</math> and <math>B</math> are on the same side of <math>CD</math>. | ||
+ | |||
+ | Now, we compute the volume of the tetrahedron <math>ABCD</math>, denoted as <math>V</math>. We have <math>V = \frac{1}{3} A \cdot AH = \frac{40 A^2}{3 \cdot 189}</math>. | ||
+ | |||
+ | Denote by <math>r</math> the inradius of the inscribed sphere in <math>ABCD</math>. | ||
+ | Denote by <math>I</math> the incenter. | ||
+ | Thus, the volume of <math>ABCD</math> can be alternatively calculated as | ||
+ | \begin{align*} | ||
+ | V & = {\rm Vol} \ IABC + {\rm Vol} \ IACD + {\rm Vol} \ IABD + {\rm Vol} \ IBCD \\ | ||
+ | & = \frac{1}{3} r \cdot 4A . | ||
+ | \end{align*} | ||
+ | |||
+ | From our two methods to compute the volume of <math>ABCD</math> and equating them, we get | ||
+ | \begin{align*} | ||
+ | r & = \frac{10A}{189} \\ | ||
+ | & = \frac{20 \sqrt{21}}{63} . | ||
+ | \end{align*} | ||
+ | |||
+ | Therefore, the answer is <math>20 + 21 + 63 = \boxed{\textbf{(104) }}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 5(A quicker method to compute the height from <math>A</math> to plane <math>BCD</math>)== | ||
+ | |||
+ | We put the solid to a 3-d coordinate system. Let <math>B = \left( 0, 0, 0 \right)</math>, <math>D = \left( \sqrt{80}, 0, 0 \right)</math>. | ||
+ | We put <math>C</math> on the <math>x-o-y</math> plane. | ||
+ | Now ,we compute the coordinates of <math>C</math>. | ||
+ | |||
+ | Applying the law of cosines on <math>\triangle BCD</math>, we get <math>\cos \angle CBD = \frac{4}{\sqrt{41 \cdot 5}}</math>. | ||
+ | Thus, <math>\sin \angle CBD = \frac{3 \sqrt{21}}{\sqrt{41 \cdot 5}}</math>. | ||
+ | Thus, <math>C = \left( \frac{4}{\sqrt{5}} , \frac{3 \sqrt{21}}{\sqrt{5}} , 0 \right)</math>. | ||
+ | |||
+ | Denote <math>A = \left( x, y , z \right)</math> with <math>z > 0</math>. | ||
+ | |||
+ | Because <math>AB = \sqrt{89}</math>, we have | ||
+ | <math>x^2 + y^2 + z^2 = 89 </math> | ||
+ | |||
+ | Because <math>AD = \sqrt{41}</math>, we have | ||
+ | \[ | ||
+ | \left( x - \sqrt{80} \right)^2 + y^2 + z^2 = 41 \hspace{1cm} (2) | ||
+ | \] | ||
+ | |||
+ | Because <math>AC = \sqrt{80}</math>, we have | ||
+ | \[ | ||
+ | \left( x - \frac{4}{\sqrt{5}} \right)^2 + \left( y - \frac{3 \sqrt{21}}{\sqrt{5}} \right)^2 | ||
+ | + z^2 = 80 \hspace{1cm} (3) | ||
+ | \] | ||
+ | |||
+ | Now, we compute <math>x</math>, <math>y</math> and <math>z</math>. | ||
+ | |||
+ | Taking <math>(1)-(2)</math>, we get | ||
+ | \[ | ||
+ | 2 \sqrt{80} x = 128 . | ||
+ | \] | ||
+ | |||
+ | Thus, <math>x = \frac{16}{\sqrt{5}}</math>. | ||
+ | |||
+ | Taking <math>(1) - (3)</math>, we get | ||
+ | \[ | ||
+ | 2 \cdot \frac{4}{\sqrt{5}} x + 2 \cdot \frac{3 \sqrt{21}}{\sqrt{5}} y | ||
+ | = 50 . | ||
+ | \] | ||
+ | |||
+ | Thus, <math>y = \frac{61}{3 \sqrt{5 \cdot 21}}</math>. | ||
+ | |||
+ | Plugging <math>x</math> and <math>y</math> into Equation (1), we get <math>z = \frac{80 \sqrt{21}}{63}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 6 (Different Perspective)== | ||
+ | Consider the following construction of the tetrahedron. Place <math>AB</math> on the floor. Construct an isosceles vertical triangle with <math>AB</math> as its base and <math>M</math> as the top vertex. Place <math>CD</math> on the top vertex parallel to the ground with midpoint <math>M.</math> Observe that <math>CD</math> can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project <math>AB</math> onto the plane of <math>CD</math>, let the minor angle <math>\theta</math> be this discrepancy. | ||
+ | |||
+ | By Median formula or Stewart's theorem, <math>AM = \frac{1}{2}\sqrt{2AC^2 + 2AD^2 - CD^2} = \frac{3\sqrt{33}}{2}.</math> Consequently the area of <math>\triangle AMB</math> is <math>\frac{\sqrt{41}}{2} \left (\sqrt{(\frac{3\sqrt{33}}{2})^2 - (\frac{\sqrt{41}}{2})^2} \right ) = 4\sqrt{41}.</math> Note the altitude <math>8</math> is also the distance between the parallel planes containing <math>AB</math> and <math>CD.</math> | ||
+ | |||
+ | By Distance Formula, | ||
+ | <cmath>\begin{align*} | ||
+ | (\frac{\sqrt{41}}{2} - \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AC^2 = 80 \\ | ||
+ | (\frac{\sqrt{41}}{2} + \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AD^2 = 89 \\ | ||
+ | \implies CD \cos{\theta} \sqrt{41} &= 9 \\ | ||
+ | \sin{\theta} &= \sqrt{1 - (\frac{9}{41})^2} = \frac{40}{41}. | ||
+ | \end{align*}</cmath> | ||
+ | Then the volume of the tetrahedron is given by <math>\frac{1}{3} [AMB] \cdot CD \sin{\theta} = \frac{160}{3}.</math> | ||
+ | |||
+ | The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from <math>I</math> w.r.t each of the faces. If <math>r</math> is the inradius, i.e the distance to the faces, then <math>\frac{1}{3} r([ABC] + [ABD] + [ACD] + [BCD])</math> must the volume. Each face has the same area by SSS congruence, and by Heron's it is <math>\frac{1}{4}\sqrt{(a + b + c)(a + b - c)(c + (a-b))(c -(a - b))} = 6\sqrt{21}.</math> | ||
+ | |||
+ | Therefore the answer is, <math>\dfrac{3 \frac{160}{3}}{24 \sqrt{21}} = \frac{20\sqrt{21}}{63} \implies \boxed{104}.</math> | ||
+ | |||
+ | ~Aaryabhatta1 | ||
+ | |||
+ | ==Solution 7 (simplest way to calculate [ABCD])== | ||
+ | After finding that <math>ABCD</math> can be inscribed in a <math>4\times5\times8</math> box, note that <math>[ABCD]=\mathbf{box}-\frac{4}{3}(4)(5)(8)=\frac{160}{3}</math>, where the third term describes the area of 4 congruent right tetrahedrons whose right angle is at the apex and whose apex lengths are 4, 5, and 8. Then proceed as in Solution 3. | ||
+ | |||
+ | ~clarkculus | ||
+ | |||
+ | ==Solution 8 (fast)== | ||
+ | Observe that <math>ABCD</math> can be inscribed in a <math>4\times5\times8</math> box, which gives the coordinates of the points: we have <math>A=(8,0,0)</math>, <math>B=(8,5,4)</math>, <math>C=(0,0,4)</math> and <math>D=(0,5,0)</math>. Recall the formula <math>V=\frac{1}{3}RS</math>, where <math>V</math> is the volume of the tetrahedron, <math>S</math> is the surface area, and <math>R</math> is the insphere radius. Using the fact that the faces of <math>ABCD</math> have equal area and that <math>V=\frac{1}{3}[ACD] \times h</math>, where <math>h</math> is the distance from plane <math>ACD</math> to <math>B</math>, we can rearrange to get <math>R=\frac{h}{4}</math>. We compute that the the equation of plane <math>ACD</math> is <math>5x+8y+10z-40=0</math>, and applying point-to-plane gives <math>h=\frac{\vert 5(8)+8(5)+10(4)-40 \vert}{\sqrt{5^2+8^2+10^2}}=\frac{80}{\sqrt{189}} \implies R=\frac{20\sqrt{21}}{63} \implies \boxed{104}</math>. | ||
+ | |||
+ | ~bomberdoodles | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/tq6lraC5prQ?si=5q1NX80POeR949qs | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution 1 by OmegaLearn.org== | ||
+ | https://youtu.be/qtu0HTFCsqc | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | |||
+ | https://youtu.be/1ZtLfJ77Ycg | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== | ||
{{AIME box|year=2024|n=I|num-b=13|num-a=15}} | {{AIME box|year=2024|n=I|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:28, 30 July 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3(Formula Abuse)
- 5 Solution 4
- 6 Solution 5(A quicker method to compute the height from to plane )
- 7 Solution 6 (Different Perspective)
- 8 Solution 7 (simplest way to calculate [ABCD])
- 9 Solution 8 (fast)
- 10 Video Solution
- 11 Video Solution 1 by OmegaLearn.org
- 12 Video Solution 2
- 13 See also
Problem
Let be a tetrahedron such that
,
, and
. There exists a point
inside the tetrahedron such that the distances from
to each of the faces of the tetrahedron are all equal. This distance can be written in the form
, where
,
, and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
.
Solution 1
Notice that \(41=4^2+5^2\), \(89=5^2+8^2\), and \(80=8^2+4^2\), let \(A~(0,0,0)\), \(B~(4,5,0)\), \(C~(0,5,8)\), and \(D~(4,0,8)\). Then the plane \(BCD\) has a normal \begin{equation*} \mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}. \end{equation*} Hence, the distance from \(A\) to plane \(BCD\), or the height of the tetrahedron, is \begin{equation*} h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}. \end{equation*} Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it \(S\). Then by the volume formula for pyramids, \begin{align*} \frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\ &=\frac13Sr\cdot4. \end{align*} Hence, \(r=\tfrac h4=\tfrac{20\sqrt{21}}{63}\), and so the answer is \(20+21+63=\boxed{104}\).
Solution by Quantum-Phantom
Solution 2
Inscribe tetrahedron in an rectangular prism as shown above.
By the Pythagorean theorem, we note
Solving yields and
Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of
We know that the distance from all faces must be the same, so we only need to find the distance from the center to plane
.
Let and
We obtain that the plane of
can be marked as
or
and the center of the prism is
Using the Point-to-Plane distance formula, our distance is
Our answer is
Solution 3(Formula Abuse)
We use the formula for the volume of iscoceles tetrahedron.
Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find .
From this, we find and can find the area of
as
Let be the distance we want to find. By taking the sum of (equal) volumes
We have
Plugging in and simplifying, we get
for an answer of
~AtharvNaphade
Solution 4
Let be perpendicular to
that meets this plane at point
.
Let
,
, and
be heights to lines
,
, and
with feet
,
, and
, respectively.
We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as , is
.
Hence, by using this area, we can compute ,
and
.
We have
,
, and
.
Because , we have
. Recall that
.
Hence,
. Hence,
.
Analogously, and
.
We introduce a function for
that is equal to 1 (resp. -1) if point
and the opposite vertex of side
are on the same side (resp. opposite sides) of side
.
The area of is
\begin{align*}
A & = \epsilon_{BC} {\rm Area} \ \triangle HBC
+ \epsilon_{CD} {\rm Area} \ \triangle HCD
+ \epsilon_{BD} {\rm Area} \ \triangle HBD \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot HP
+ \frac{1}{2} \epsilon_{CD} CD \cdot HQ +
\frac{1}{2} \epsilon_{BD} BD \cdot HR \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot \sqrt{AP^2 - AH^2}
+ \frac{1}{2} \epsilon_{CD} CD \cdot \sqrt{AQ^2 - AH^2} \\
& \quad + \frac{1}{2} \epsilon_{BD} CD \cdot \sqrt{AR^2 - AH^2} . \hspace{1cm} (1)
\end{align*}
Denote .
The above equation can be organized as
\begin{align*}
B & = \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
+ \epsilon_{CD} \sqrt{B^2 - 41 AH^2} \\
& \quad + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
\end{align*}
This can be further reorganized as \begin{align*} B - \epsilon_{BC} \sqrt{B^2 - 89 AH^2} & = \epsilon_{CD} \sqrt{B^2 - 41 AH^2} + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} . \end{align*}
Taking squares on both sides and reorganizing terms, we get \begin{align*} & 16 AH^2 - \epsilon_{BC} B \sqrt{B^2 - 89 AH^2} \\ & = \epsilon_{CD} \epsilon_{BD} \sqrt{\left( B^2 - 41 AH^2 \right) \left( B^2 - 80 AH^2 \right)} . \end{align*}
Taking squares on both sides and reorganizing terms, we get
Taking squares on both sides, we finally get \begin{align*} AH & = \frac{20B}{189} \\ & = \frac{40A}{189}. \end{align*}
Now, we plug this solution to Equation (1). We can see that ,
.
This indicates that
is out of
. To be specific,
and
are on opposite sides of
,
and
are on the same side of
, and
and
are on the same side of
.
Now, we compute the volume of the tetrahedron , denoted as
. We have
.
Denote by the inradius of the inscribed sphere in
.
Denote by
the incenter.
Thus, the volume of
can be alternatively calculated as
\begin{align*}
V & = {\rm Vol} \ IABC + {\rm Vol} \ IACD + {\rm Vol} \ IABD + {\rm Vol} \ IBCD \\
& = \frac{1}{3} r \cdot 4A .
\end{align*}
From our two methods to compute the volume of and equating them, we get
\begin{align*}
r & = \frac{10A}{189} \\
& = \frac{20 \sqrt{21}}{63} .
\end{align*}
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 5(A quicker method to compute the height from
to plane
)
We put the solid to a 3-d coordinate system. Let ,
.
We put
on the
plane.
Now ,we compute the coordinates of
.
Applying the law of cosines on , we get
.
Thus,
.
Thus,
.
Denote with
.
Because , we have
Because , we have
\[
\left( x - \sqrt{80} \right)^2 + y^2 + z^2 = 41 \hspace{1cm} (2)
\]
Because , we have
\[
\left( x - \frac{4}{\sqrt{5}} \right)^2 + \left( y - \frac{3 \sqrt{21}}{\sqrt{5}} \right)^2
+ z^2 = 80 \hspace{1cm} (3)
\]
Now, we compute ,
and
.
Taking , we get
\[
2 \sqrt{80} x = 128 .
\]
Thus, .
Taking , we get
\[
2 \cdot \frac{4}{\sqrt{5}} x + 2 \cdot \frac{3 \sqrt{21}}{\sqrt{5}} y
= 50 .
\]
Thus, .
Plugging and
into Equation (1), we get
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 6 (Different Perspective)
Consider the following construction of the tetrahedron. Place on the floor. Construct an isosceles vertical triangle with
as its base and
as the top vertex. Place
on the top vertex parallel to the ground with midpoint
Observe that
can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project
onto the plane of
, let the minor angle
be this discrepancy.
By Median formula or Stewart's theorem, Consequently the area of
is
Note the altitude
is also the distance between the parallel planes containing
and
By Distance Formula,
Then the volume of the tetrahedron is given by
The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from w.r.t each of the faces. If
is the inradius, i.e the distance to the faces, then
must the volume. Each face has the same area by SSS congruence, and by Heron's it is
Therefore the answer is,
~Aaryabhatta1
Solution 7 (simplest way to calculate [ABCD])
After finding that can be inscribed in a
box, note that
, where the third term describes the area of 4 congruent right tetrahedrons whose right angle is at the apex and whose apex lengths are 4, 5, and 8. Then proceed as in Solution 3.
~clarkculus
Solution 8 (fast)
Observe that can be inscribed in a
box, which gives the coordinates of the points: we have
,
,
and
. Recall the formula
, where
is the volume of the tetrahedron,
is the surface area, and
is the insphere radius. Using the fact that the faces of
have equal area and that
, where
is the distance from plane
to
, we can rearrange to get
. We compute that the the equation of plane
is
, and applying point-to-plane gives
.
~bomberdoodles
Video Solution
https://youtu.be/tq6lraC5prQ?si=5q1NX80POeR949qs
~MathProblemSolvingSkills.com
Video Solution 1 by OmegaLearn.org
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
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