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

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==Solution==
 
==Solution==
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Assume <math>CD > 1</math> and assume that all other sides are <math>\ le 1</math>.
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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
 
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>,
 
<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>,
Line 38: Line 38:
 
[[File:Prob_1967_2_fig1.png|600px]]
 
[[File:Prob_1967_2_fig1.png|600px]]
  
−
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>.
+
At least one of the segments <math>AP, PB</math> has to be <math>\ge \frac{x}{2}</math>.
 +
Suppose <math>PB \ge \frac{x}{2}</math>.  (If <math>AP</math> were bigger that <math>\frac{x}{2}</math>
 +
the argument would be the same.)  We have that
 +
<math>CP = \sqrt{BC^2 - PB^2} \le \sqrt{1 - \frac{x^2}{4}}</math>.  By the same
 +
argument in <math>\triangle ABD</math> we have <math>DR \le \sqrt{1 - \frac{x^2}{4}}</math>.
 +
Since <math>CQ \perp</math> plane <math>ABD</math>, we have <math>CQ \le CP</math>, so
 +
<math>CQ \le \sqrt{1 - \frac{x^2}{4}}</math>.
 +
 +
The volume of the tetrahedron is
  
−
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>.
+
<math>V = \frac{1}{3} \cdot (</math>area of <math>\triangle ABD) \cdot</math> height from <math>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)</math>.
  
 +
We need to prove that <math>\frac{x}{6} \left( 1 - \frac{x^2}{4} \right) \le \frac{1}{8}</math>.
 +
Some simple computations show that this is the same as <math>(1 - x)(3 - x - x^2) \ge 0</math>.
 +
This is true because <math>0 < x \le 1</math>, and <math>x^2 - x + 3 \ge 0</math> on this interval.
  
 +
Note: <math>V = \frac{1}{8}</math> is achieved when <math>x = 1</math> and all inequalities
 +
are equalities.  This is the case when all sides except <math>AD</math> are <math>= 1</math>,
 +
<math>P, R</math> are midpoints of <math>AB</math> and <math>Q = P</math> (in which case the planes
 +
<math>ABC, ABD</math> are perpendicular).  In this case, <math>AD = \frac{\sqrt{6}}{2}</math>,
 +
as can be seen from an easy computation.
  
  
 +
==Solution 2==
  
−
TO BE CONTINUED.  dOING A SAVE MIDWAY SP I DON'T LOOSE WORK DONE SO FAR.
+
 
 +
 
 +
 
 +
 
 +
 
 +
TO BE CONTINUED.  DOING A SAVE MIDWAY SO I DON'T LOOSE WORK DONE SO FAR.
  
  

Revision as of 20:20, 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

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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