View source for 1984 AIME Problems/Problem 9

== Problem ==
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 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;
}

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>
~ kilobyte144

(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.

== Solution 4 (coord/vector bash)==

We can use 3D coordinates.

Let <math>A = (0, 0, 0)</math> and <math>B = (3, 0, 0).</math> WLOG, let <math>D = \left(\frac{3}{2}, 8, 0\right)</math>, because the area of <math>\Delta{ABD} = 12</math> and the tetrahedron area won't change if we put it somewhere else with <math>y=8.</math>

To find <math>C</math>, we can again let the <math>x</math>-coordinate be <math>\frac{3}{2}</math> for simplicity. Note that <math>C</math> is <math>10</math> units away from <math>AB</math> because the area of <math>\Delta{ABC}</math> is <math>15</math>. Since the angle between <math>ABD</math> and <math>ABC</math> is <math>30^\circ</math>, we can form a 30-60-90 triangle between <math>A</math>, <math>B</math>, and an altitude dropped from <math>C</math> onto face <math>ABD</math>. Since <math>10</math> is the hypotenuse, we get <math>5\sqrt{3}</math> and <math>5</math> as legs. Then <math>y=5\sqrt{3}</math> and <math>z=5</math>, so <math>C = \left(\frac{3}{2}, 5\sqrt{3}, 5\right).</math>

(I highly advise you to draw both the tetrahedron and 30-60-90 triangle to get a better perspective.)

Now, we can move onto vectors. To find the volume of the tetrahedron, we use the formula <math>\frac{1}{3}Bh.</math> Letting <math>\Delta{ABC}</math> be the base we have <math>B = 15</math> (from the problem statement). We need to find the distance between <math>D</math> and <math>ABC</math>, and to do this, we should find the projection of <math>D</math> onto face <math>ABC</math>.

Note that we can simplify this to projecting <math>D</math> onto <math>\mathbf{\overrightarrow{C}}.</math> This is because we know the projection will have the same <math>x</math>-coordinate as <math>D</math> and <math>C</math>, as both are <math>\frac{3}{2}.</math> Now we find <math>\text{proj}_{\mathbf{\overrightarrow{D}}} \mathbf{\overrightarrow{C}}</math>, or plugging in our coordinates, <math>\text{proj}_{\langle\frac{3}{2}, 5\sqrt{3}, 5\rangle} \left\langle\frac{3}{2}, 8, 0\right\rangle</math>. 

Let the <math>x</math>-coordinates for both be <math>0</math> for simplicity, because we can always add a <math>\frac{3}{2}</math> at the end. Using the projection formula, we get <cmath>\langle 0, 6, 2\sqrt{3}\rangle.</cmath>

Finally, we calculate the distance between <math>\left(\frac{3}{2}, 6, 2\sqrt{3}\right)</math> and <math>D</math> to be <math>4</math>. So the height is <math>4</math>, and plugging into our tetrahedron formula we get <cmath>\frac{1}{3}\cdot 15\cdot 4 = \boxed{20}.</cmath>

-kilobyte144

== See also ==
{{AIME box|year=1984|num-b=8|num-a=10}}

[[Category:Intermediate Geometry Problems]]
[[Category:3D Asymptote]]