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Difference between revisions of "1967 IMO Problems/Problem 2"

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We begin with a simple observation.  Let <math>ABCD</math> be a tetrahedron, and
 
We begin with a simple observation.  Let <math>ABCD</math> be a tetrahedron, and
 
consider the transformations which rotate <math>\triangle ABC</math> around <math>AB</math>
 
consider the transformations which rotate <math>\triangle ABC</math> around <math>AB</math>
−
as an axis while keeping <math>\triangle ABD</math> fixed.  (The lengths of all
+
while keeping <math>\triangle ABD</math> fixed.  We get a set of tetrahedrons,
−
sides except <math>CD</math> are constant through this transformation.)
+
two of which, <math>ABC_1D</math> and <math>ABC_2D</math> are shown in the picture below.
 +
The lengths of all sides except <math>CD</math> are constant through this
 +
transformation.
  
 
[[File:Prob_1967_2_fig2.png|600px]]
 
[[File:Prob_1967_2_fig2.png|600px]]
 +
 +
1. Assume that the angles between the planes <math>ABD</math> and <math>ABC</math>, and <math>ABD</math>
 +
and <math>ABC_1</math> are both acute.  If the perpendicular from <math>C_1</math> to the
 +
plane <math>ABD</math> is larger that the perpendicular from <math>C</math> to the plane
 +
<math>ABD</math> then the volume of <math>ABC_1D</math> is larger than the volume of <math>ABCD</math>.
 +
 +
2. Furthermore, the tetrahedron <math>ABC_2D</math> obtained when the position of
 +
<math>C_2</math> is such that the planes <math>ABD</math> and <math>ABC_2</math> are perpendicular has
 +
the maximum volume of all tetrahedrons obtained from rotating
 +
<math>\triangle ABC</math> around <math>AB</math>.
 +
  
  

Revision as of 19:59, 13 September 2024

Prove that if one and only one edge of a tetrahedron is greater than $1$, then its volume is $\le \frac{1}{8}$.

Solution

Assume $CD>1$ and let $AB=x$. Let $P,Q,R$ be the feet of perpendicular from $C$ to $AB$ and $\triangle ABD$ and from $D$ to $AB$, respectively.

Suppose $BP>PA$. We have that $CP=\sqrt{CB^2-BT^2}\le\sqrt{1-\frac{x^2}4}$, $CQ\le CP\le\sqrt{1-\frac{x^2}4}$. We also have $DQ^2\le\sqrt{1-\frac{x^2}4}$. So the volume of the tetrahedron is $\frac13\left(\frac12\cdot AB\cdot DR\right)CQ\le\frac{x}6\left(1-\frac{x^2}4\right)$.

We want to prove that this value is at most $\frac18$, which is equivalent to $(1-x)(3-x-x^2)\ge0$. This is true because $0<x\le 1$.

The above solution was posted and copyrighted by jgnr. The original thread can be found here: [1]


Remarks (added by pf02, September 2024)

The solution above is essentially correct, and it is nice, but it is so sloppily written that it borders the incomprehensible. Below I will give an edited version of it for the sake of completeness.

Then, I will give a second solution to the problem.

A few notes which may be of interest.

The condition that one side is greater than $1$ is not really necessary. The statement is true even if all sides are $\le 1$. What we need is that no more than one side is $> 1$.

The upper limit of $1/8$ for the volume of the tetrahedron is actually reached. This will become clear from both solutions.


Solution

Assume $CD > 1$ and assume that all other sides are $\le 1$. Let $AB = x$. Let $P, Q, R$ be the feet of perpendiculars from $C$ to $AB$, from $C$ to the plane $ABD$, and from $D$ to $AB$, respectively.

Prob 1967 2 fig1.png

At least one of the segments $AP, PB$ has to be $\ge \frac{x}{2}$. Suppose $PB \ge \frac{x}{2}$. (If $AP$ were bigger that $\frac{x}{2}$ the argument would be the same.) We have that $CP = \sqrt{BC^2 - PB^2} \le \sqrt{1 - \frac{x^2}{4}}$. By the same argument in $\triangle ABD$ we have $DR \le \sqrt{1 - \frac{x^2}{4}}$. Since $CQ \perp$ plane $ABD$, we have $CQ \le CP$, so $CQ \le \sqrt{1 - \frac{x^2}{4}}$.

The volume of the tetrahedron is

$V = \frac{1}{3} \cdot ($area of $\triangle ABD) \cdot$ height from $C = \frac{1}{3} \cdot \left( \frac{1}{2} \cdot AB \cdot DR \right) \cdot CQ \le \left( \frac{1}{6} \cdot x \cdot \sqrt{1 - \frac{x^2}{4}} \cdot \sqrt{1 - \frac{x^2}{4}} \right) = \frac{x}{6} \left( 1 - \frac{x^2}{4} \right)$.

We need to prove that $\frac{x}{6} \left( 1 - \frac{x^2}{4} \right) \le \frac{1}{8}$. Some simple computations show that this is the same as $(1 - x)(3 - x - x^2) \ge 0$. This is true because $0 < x \le 1$, and $-x^2 - x + 3 \ge 0$ on this interval.

Note: $V = \frac{1}{8}$ is achieved when $x = 1$ and all inequalities are equalities. This is the case when all sides except $AD$ are $= 1$, $P, R$ are midpoints of $AB$ and $Q = P$ (in which case the planes $ABC, ABD$ are perpendicular). In this case, $AD = \frac{\sqrt{6}}{2}$, as can be seen from an easy computation.


Solution 2

We begin with a simple observation. Let $ABCD$ be a tetrahedron, and consider the transformations which rotate $\triangle ABC$ around $AB$ while keeping $\triangle ABD$ fixed. We get a set of tetrahedrons, two of which, $ABC_1D$ and $ABC_2D$ are shown in the picture below. The lengths of all sides except $CD$ are constant through this transformation.

Prob 1967 2 fig2.png

1. Assume that the angles between the planes $ABD$ and $ABC$, and $ABD$ and $ABC_1$ are both acute. If the perpendicular from $C_1$ to the plane $ABD$ is larger that the perpendicular from $C$ to the plane $ABD$ then the volume of $ABC_1D$ is larger than the volume of $ABCD$.

2. Furthermore, the tetrahedron $ABC_2D$ obtained when the position of $C_2$ is such that the planes $ABD$ and $ABC_2$ are perpendicular has the maximum volume of all tetrahedrons obtained from rotating $\triangle ABC$ around $AB$.



(Solution by pf02, September 2024)

TO BE CONTINUED. DOING A SAVE MIDWAY SO I DON'T LOOSE WORK DONE SO FAR.


See Also

1967 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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