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Difference between revisions of "1967 IMO Problems/Problem 2"
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The above solution was posted and copyrighted by jgnr. The original thread can be found here: [https://aops.com/community/p1480514]
 
The above solution was posted and copyrighted by jgnr. The original thread can be found here: [https://aops.com/community/p1480514]
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 +
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==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 <math>1</math> is not really
 +
necessary.  The statement is true even if all sides are <math>\le 1</math>.
 +
What we need is that no more than one side is <math>> 1</math>.
 +
 +
The upper limit of <math>1/8</math> for the volume of the tetrahedron
 +
is actually reached.  This will become clear from both solutions.
 +
 +
 +
==Solution==
 +
Assume <math>CD > 1</math> and assume that all other sides are <math>\ le 1</math>.
 +
Let <math>AB = x</math>.  Let <math>P, Q, R</math> be the feet of perpendiculars from
 +
<math>C</math> to <math>AB</math>, from <math>C</math> to the plane <math>ABD</math>, and from <math>D</math> to <math> AB</math>,
 +
respectively.
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[[File:Prob_1967_2_fig1.png|600px]]
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Suppose <math> BP>PA</math>. We have that <math> CP=\sqrt{CB^2-BT^2}\le\sqrt{1-\frac{x^2}4}</math>, <math> CQ\le CP\le\sqrt{1-\frac{x^2}4}</math>. We also have <math> DQ^2\le\sqrt{1-\frac{x^2}4}</math>. So the volume of the tetrahedron is <math> \frac13\left(\frac12\cdot AB\cdot DR\right)CQ\le\frac{x}6\left(1-\frac{x^2}4\right)</math>.
 +
 +
We want to prove that this value is at most <math> \frac18</math>, which is equivalent to <math> (1-x)(3-x-x^2)\ge0</math>. This is true because <math> 0<x\le 1</math>.
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TO BE CONTINUED.  dOING A SAVE MIDWAY SP I DON'T LOOSE WORK DONE SO FAR.
  
  

Revision as of 18:10, 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

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 $0<x\le 1$.



TO BE CONTINUED. dOING A SAVE MIDWAY SP 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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