Difference between revisions of "1984 AIME Problems/Problem 9"
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== Solution 2 (Rigorous)== | == 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*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>bh/3=15*4/3=\boxed{020}</math> | 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*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>bh/3=15*4/3=\boxed{020}</math> | ||
| + | ~ ShreyJ | ||
== Solution 3 Kinda Sketchy== | == Solution 3 Kinda Sketchy== | ||
Revision as of 15:05, 18 June 2018
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
![[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; } 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]](http://latex.artofproblemsolving.com/c/e/a/ceabcb49396ab28000e1343ca6499fb8ae910ebb.png)
Position face on the bottom. Since ![$[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}$](http://latex.artofproblemsolving.com/e/1/4/e144a1a6fbdb264ee985bc04ba58ac14ab8e8d47.png)
, 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 ![$[DXC]=\frac{ab\sin{c}}{2}=20=5*h \rightarrow h = 4$](http://latex.artofproblemsolving.com/6/c/1/6c1d78574b909eef0aedcda17d6ef68fdc7d2838.png)
where h is the height of the tetrahedron from . Hence, the volume of the tetrahedron is 
~ ShreyJ
Solution 3 Kinda 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 | ||
