Difference between revisions of "1984 AIME Problems/Problem 9"
(replace with 3d asymptote) |
m (compatible with three?) |
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== Solution == | == Solution == | ||
<span style="font-size:50%">For non-asymptote version of image, see [[:Image:1984_AIME-9.png]].</span><center><asy> | <span style="font-size:50%">For non-asymptote version of image, see [[:Image:1984_AIME-9.png]].</span><center><asy> | ||
+ | /* modified version of olympiad modules */ | ||
+ | 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); | size(200); | ||
− | import three; | + | 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); | 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); | 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 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 */ | /* 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> | </asy></center> | ||
Line 24: | Line 70: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Asymptote]] |
Revision as of 15:18, 16 March 2010
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
For non-asymptote version of image, see Image:1984_AIME-9.png.
/* modified version of olympiad modules */ 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); (Error making remote request. Unknown error_msg)
Position face on the bottom. Since , we find that . The height of forms a with the height of the tetrahedron, so . The volume of the tetrahedron is thus .
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 |