Difference between revisions of "1984 AIME Problems/Problem 9"
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== Solution 3 (Sketchy)== | == 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. | 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> | ||
+ | |||
+ | -PureSwag | ||
== See also == | == See also == |
Latest revision as of 20:44, 26 November 2021
Contents
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
, 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.
Solution 4 (coord/vector bash)
We can use 3D coordinates.
Let and
WLOG, let
, because the area of
and the tetrahedron area won't change if we put it somewhere else with
To find , we can again let the
-coordinate be
for simplicity. Note that
is
units away from
because the area of
is
. Since the angle between
and
is
, we can form a 30-60-90 triangle between
,
, and an altitude dropped from
onto face
. Since
is the hypotenuse, we get
and
as legs. Then
and
, so
(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 Letting
be the base we have
(from the problem statement). We need to find the distance between
and
, and to do this, we should find the projection of
onto face
.
Note that we can simplify this to projecting onto
This is because we know the projection will have the same
-coordinate as
and
, as both are
Now we find
, or plugging in our coordinates,
.
Let the -coordinates for both be
for simplicity, because we can always add a
at the end. Using the projection formula, we get
Finally, we calculate the distance between and
to be
. So the height is
, and plugging into our tetrahedron formula we get
-PureSwag
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 |