Difference between revisions of "1984 AIME Problems/Problem 9"
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In [[tetrahedron]] <math>ABCD</math>, [[edge]] <math>AB</math> has length 3 cm. The area of [[face]] <math>ABC</math> is <math>15\mbox{cm}^2</math> and the area of face <math>ABD</math> is <math>12 \mbox { cm}^2</math>. These two faces meet each other at a <math>30^\circ</math> angle. Find the [[volume]] of the tetrahedron in <math>\mbox{cm}^3</math>. | In [[tetrahedron]] <math>ABCD</math>, [[edge]] <math>AB</math> has length 3 cm. The area of [[face]] <math>ABC</math> is <math>15\mbox{cm}^2</math> and the area of face <math>ABD</math> is <math>12 \mbox { cm}^2</math>. These two faces meet each other at a <math>30^\circ</math> angle. Find the [[volume]] of the tetrahedron in <math>\mbox{cm}^3</math>. | ||
− | == Solution == | + | == Solution 1== |
− | [[ | + | <center><asy> |
+ | /* modified version of olympiad modules */ | ||
+ | import three; | ||
+ | real markscalefactor = 0.03; | ||
+ | path3 rightanglemark(triple A, triple B, triple C, real s=8) | ||
+ | { | ||
+ | triple P,Q,R; | ||
+ | P=s*markscalefactor*unit(A-B)+B; | ||
+ | R=s*markscalefactor*unit(C-B)+B; | ||
+ | Q=P+R-B; | ||
+ | return P--Q--R; | ||
+ | } | ||
+ | path3 anglemark(triple A, triple B, triple C, real t=8 ... real[] s) | ||
+ | { | ||
+ | triple M,N,P[],Q[]; | ||
+ | path3 mark; | ||
+ | int n=s.length; | ||
+ | M=t*markscalefactor*unit(A-B)+B; | ||
+ | N=t*markscalefactor*unit(C-B)+B; | ||
+ | for (int i=0; i<n; ++i) | ||
+ | { | ||
+ | P[i]=s[i]*markscalefactor*unit(A-B)+B; | ||
+ | Q[i]=s[i]*markscalefactor*unit(C-B)+B; | ||
+ | } | ||
+ | mark=arc(B,M,N); | ||
+ | for (int i=0; i<n; ++i) | ||
+ | { | ||
+ | if (i%2==0) | ||
+ | { | ||
+ | mark=mark--reverse(arc(B,P[i],Q[i])); | ||
+ | } | ||
+ | else | ||
+ | { | ||
+ | mark=mark--arc(B,P[i],Q[i]); | ||
+ | } | ||
+ | } | ||
+ | if (n%2==0 && n!=0) | ||
+ | mark=(mark--B--P[n-1]); | ||
+ | else if (n!=0) | ||
+ | mark=(mark--B--Q[n-1]); | ||
+ | else mark=(mark--B--cycle); | ||
+ | return mark; | ||
+ | } | ||
− | Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. | + | size(200); |
+ | import three; defaultpen(black+linewidth(0.7)); pen small = fontsize(10); | ||
+ | triple A=(0,0,0),B=(3,0,0),C=(1.8,10,0),D=(1.5,4,4),Da=(D.x,D.y,0),Db=(D.x,0,0); | ||
+ | currentprojection=perspective(16,-10,8); | ||
+ | |||
+ | draw(surface(A--B--C--cycle),rgb(0.6,0.7,0.6),nolight); | ||
+ | draw(surface(A--B--D--cycle),rgb(0.7,0.6,0.6),nolight); | ||
+ | |||
+ | /* draw pyramid - other lines + angles */ | ||
+ | draw(A--B--C--A--D--B--D--C); | ||
+ | draw(D--Da--Db--cycle); | ||
+ | draw(rightanglemark(D,Da,Db));draw(rightanglemark(A,Db,D));draw(anglemark(Da,Db,D,15)); | ||
+ | |||
+ | /* labeling points */ | ||
+ | label("$A$",A,SW);label("$B$",B,S);label("$C$",C,S);label("$D$",D,N);label("$30^{\circ}$",Db+(0,.35,0.08),(1.5,1.2),small); | ||
+ | label("$3$",(A+B)/2,S); label("$15\mathrm{cm}^2$",(Db+C)/2+(0,-0.5,-0.1),NE,small); label("$12\mathrm{cm}^2$",(A+D)/2,NW,small); | ||
+ | </asy></center> | ||
+ | |||
+ | Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. Because the problem does not specify, we may assume both <math>ABC</math> and <math>ABD</math> to be isosceles triangles. Thus, the height of <math>ABD</math> forms a <math>30-60-90</math> with the height of the tetrahedron. So, <math>h = \frac{1}{2} (8) = 4</math>. The volume of the tetrahedron is thus <math>\frac{1}{3}Bh = \frac{1}{3} \cdot15 \cdot 4 = \boxed{020}</math>. | ||
+ | |||
+ | == Solution 2 (Rigorous)== | ||
+ | It is clear that <math>DX=8</math> and <math>CX=10</math> where <math>X</math> is the foot of the perpendicular from <math>D</math> and <math>C</math> to side <math>AB</math>. Thus <math>[DXC]=\frac{ab\sin{c}}{2}=20=5 \cdot h \rightarrow h = 4</math> where h is the height of the tetrahedron from <math>D</math>. Hence, the volume of the tetrahedron is <math>\frac{bh}{3}=15\cdot \frac{4}{3}=\boxed{020}</math> | ||
+ | ~ Mathommill | ||
+ | |||
+ | |||
+ | (Note this actually isn't rigorous because they never proved that the height from <math>D</math> to <math>XC</math> is the altitude of the tetrahedron. | ||
+ | |||
+ | == Solution 3 (Sketchy)== | ||
+ | Make faces <math>ABC</math> and <math>ABD</math> right triangles. This makes everything a lot easier. Then do everything in solution 1. | ||
== See also == | == See also == | ||
{{AIME box|year=1984|num-b=8|num-a=10}} | {{AIME box|year=1984|num-b=8|num-a=10}} | ||
− | |||
− | |||
− | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Asymptote]] |
Revision as of 21:17, 18 October 2020
Problem
In tetrahedron , edge has length 3 cm. The area of face is and the area of face is . These two faces meet each other at a angle. Find the volume of the tetrahedron in .
Solution 1
Position face on the bottom. Since , we find that . Because the problem does not specify, we may assume both and to be isosceles triangles. Thus, the height of forms a with the height of the tetrahedron. So, . The volume of the tetrahedron is thus .
Solution 2 (Rigorous)
It is clear that and where is the foot of the perpendicular from and to side . Thus where h is the height of the tetrahedron from . Hence, the volume of the tetrahedron is ~ Mathommill
(Note this actually isn't rigorous because they never proved that the height from to is the altitude of the tetrahedron.
Solution 3 (Sketchy)
Make faces and right triangles. This makes everything a lot easier. Then do everything in solution 1.
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |