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

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Prove that iff. one edge of a tetrahedron is less than <math>1</math>; then its volume is less than or equal to <math>\frac{1}{8}</math>.
+
==Problem==
  
−
<math>\textbf{Solution:}</math> It can be found here [https://artofproblemsolving.com/community/c6h21139p137291].
+
Prove that if one and only one edge of a tetrahedron is greater than <math>1</math>,
 +
then its volume is <math>\le \frac{1}{8}</math>.
  
 +
==Solution==
 +
 +
Assume <math> CD>1</math> and let <math> AB=x</math>. Let <math> P,Q,R</math> be the feet of perpendicular from <math> C</math> to <math> AB</math> and <math> \triangle ABD</math> and from <math> D</math> to <math> AB</math>, respectively.
 +
 +
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>.
 +
 +
The above solution was posted and copyrighted by jgnr. The original thread can be found here: [https://aops.com/community/p1480514]
 +
 +
 +
==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 that five of the edges are <math>\le 1</math>.  Take them to be the
 +
edges other than <math>CD</math>.  Denote <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.
 +
 +
[[File:Prob_1967_2_fig1.png|600px]]
 +
 +
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 <math>V</math> of the tetrahedron is
 +
 +
<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 > 0</math> on the interval <math>(0, 1]</math>.
 +
 +
===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>CD</math> are <math>= 1</math>,
 +
<math>P = R</math> are the midpoint of <math>AB</math>, and <math>Q = P</math> (in which case the planes
 +
<math>ABC, ABD</math> are perpendicular).  In this case, <math>CD = \frac{\sqrt{6}}{2}</math>,
 +
and <math>V = \frac{1}{8}</math> as can be seen from an easy computation.
 +
 +
[This is an edited version of the solution by jgnr.]
 +
 +
 +
==Solution 2==
 +
 +
Let <math>\mathcal{T}</math> be the set of tetrahedrons with five edges <math>\le 1</math>.
 +
This proof will show that there is a <math>T \in \mathcal{T}</math> with one edge
 +
<math>> 1</math> and such that <math>\mathbf{volume} (T) = \frac{1}{8}</math>, and that for
 +
any <math>U \in \mathcal{T}</math> either <math>U = T</math> or there is a finite sequence
 +
of tetrahedrons <math>T_1, \dots, T_n</math> such that
 +
 +
<math>\mathbf{volume} (U) = \mathbf{volume} (T_1) < \dots <
 +
\mathbf{volume} (T_n) = \mathbf{volume} (T)</math>.
 +
 +
The statement of the problem is a consequence of these facts.
 +
 +
We begin with two simple propositions.
 +
 +
===Proposition===
 +
 +
Let <math>ABCD</math> be a tetrahedron, and consider the transformations which
 +
rotate <math>\triangle ABC</math> around <math>AB</math> while keeping <math>\triangle ABD</math>
 +
fixed.  We get a set of tetrahedrons, 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]]
 +
 +
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>.
 +
 +
These statements are intuitively clear, since the volume <math>V</math> of the
 +
tetrahedron <math>ABCD</math> is given by
 +
 +
<math>V = \frac{1}{3} \cdot (</math>area of <math>\triangle ABD) \cdot (</math>height from <math>C)</math>.
 +
 +
A formal proof is very easy, and I will skip it.
 +
 +
===Corollary===
 +
 +
Given a tetrahedron <math>T</math>, and an edge <math>e_1</math> of it, we can find another
 +
tetrahedron <math>U</math> such that <math>\mathbf{volume}(U) > \mathbf{volume}(T)</math>,
 +
with an edge <math>f_1 > e_1</math>, and such that all the other edges of <math>U</math>
 +
are equal to the corresponding edges of <math>T</math>, <math>\mathbf{unless}</math> the
 +
edge <math>e_1</math> stretches between sides of <math>T</math> which are perpendicular.
 +
When we chose a bigger <math>f_1</math>, if <math>e_1 < 1</math> we can choose <math>f_1 = 1</math>.
 +
Or, we can choose <math>f_1</math> such that it stretches between sides which
 +
are perpendicular.
 +
 +
(By "stretches between two sides" I mean that the end points of the
 +
edge are the vertices on the two sides which are not common to the
 +
two sides.  In the picture above, <math>DC_2</math> stretches between the sides
 +
<math>ABD, AC_2B</math> of <math>ABC_2D</math>.)
 +
 +
===Lemma===
 +
 +
Assume we have a tetrahedron <math>T</math> with edges <math>e_1, \dots, e_6</math>, such
 +
that <math>e_2, \dots, e_6 \le 1</math>.  If there is an edge <math>e_m < 1</math> among
 +
<math>e_2, \dots, e_6</math> then there is a tetrahedron <math>U</math> with volume bigger
 +
than the volume of <math>T</math>, whose edges are equal to those of <math>T</math>,
 +
except for <math>e_m</math>, which is replaced by an edge of size <math>1</math>.
 +
 +
====Proof====
 +
 +
Case 1: If <math>T</math> does not have any sides which are perpendicular, then
 +
the existence of <math>U</math> follows from the corollary.
 +
 +
Case 2: Assume <math>T</math> has exactly two sides which are perpendicular
 +
(like <math>ABC_2D</math> in the picture above).  If <math>C_2D < 1</math> and it were
 +
the only edge <math>< 1</math>, then all the other edges are <math>= 1</math> (because
 +
they were assumed to be <math>\le 1</math>).  In this case
 +
<math>\triangle ABD, \triangle AC_2B</math> are equilateral with sides <math>= 1</math>,
 +
and the planes can not be perpendicular since <math>C_2D < 1</math>.  (Indeed,
 +
an easy computation shows that if we take two equilateral triangles
 +
<math>\triangle ABD, \triangle AC_2B</math> and place them perpendicular to each
 +
other, then <math>CD = \frac{\sqrt{6}}{2} > 1</math>.)  So <math>e_m</math> must be one of
 +
the other sides.  Then again, the existence of <math>U</math> follows from the
 +
corollary. 
 +
 +
Case 3: Assume that three sides are perpendicular.
 +
 +
[[File:Prob_1967_2_fig3.png|600px]]
 +
 +
Assume the perpendicular sides are the ones meeting at <math>A</math>, i.e.
 +
each pair of the planes meeting at <math>A</math> are perpendicular.  Since
 +
at least two of the edges <math>BD, BC, CD \le 1</math> it follows that
 +
<math>AB, AC, AD < 1</math> (the sides of the right angle in a right triangle
 +
are less than the hypotenuse).  Just for the sake of notation,
 +
assume <math>e_m = AB < 1</math>.  We can apply the corollary, and find a
 +
tetrahedron <math>U</math> with volume bigger than the volume of <math>T</math>, with
 +
edges equal to those of <math>T</math>, except that <math>AB</math> is replaced by an
 +
edge <math>= 1</math>.
 +
 +
Now the problem is very easy to prove.  Let <math>T</math> be a tetrahedron
 +
with edges <math>e_1, \dots, e_6</math>, such that <math>e_2, \dots, e_6 \le 1</math>.
 +
Apply the lemma as many times as necessary (up to five times),
 +
successively replacing each edge <math>< 1</math> by an edge <math>= 1</math>.  We
 +
obtain a tetrahedron <math>U</math> with five edges <math>f_2, \dots, f_6 = 1</math>,
 +
and one edge <math>f_1 = e_1</math>.  If <math>f_1</math> stretches between two
 +
perpendicular sides, we are done.  If not, apply the corollary
 +
one more time to obtain a bigger tetrahedron in which <math>f_1</math> is
 +
replaced by a larger edge which stretches between two
 +
perpendicular sides.
 +
 +
We obtain the same result as in the first solution: the largest
 +
tetrahedron is the one formed by two equilateral triangles with
 +
sides <math>= 1</math>, having one side in common, with the two planes
 +
containing the triangles perpendicular.  An easy calculation
 +
shows that the edge which is <math>> 1</math> is in fact of length
 +
<math>\frac{\sqrt{6}}{2}</math>, and the volume of this tetrahedron
 +
is <math>\frac{1}{8}</math>.
 +
 +
[Solution by pf02, September 2024]
 +
 +
 +
== See Also == {{IMO box|year=1967|num-b=1|num-a=3}}
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:3D Geometry Problems]]
 
[[Category:3D Geometry Problems]]

Latest revision as of 19:21, 10 November 2024

Problem

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 that five of the edges are $\le 1$. Take them to be the edges other than $CD$. Denote $AB = x$. 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 $V$ 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 $-x^2 - x + 3 > 0$ on the interval $(0, 1]$.

Note

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

[This is an edited version of the solution by jgnr.]


Solution 2

Let $\mathcal{T}$ be the set of tetrahedrons with five edges $\le 1$. This proof will show that there is a $T \in \mathcal{T}$ with one edge $> 1$ and such that $\mathbf{volume} (T) = \frac{1}{8}$, and that for any $U \in \mathcal{T}$ either $U = T$ or there is a finite sequence of tetrahedrons $T_1, \dots, T_n$ such that

$\mathbf{volume} (U) = \mathbf{volume} (T_1) < \dots < \mathbf{volume} (T_n) = \mathbf{volume} (T)$.

The statement of the problem is a consequence of these facts.

We begin with two simple propositions.

Proposition

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 $AB$.

These statements are intuitively clear, since the volume $V$ of the tetrahedron $ABCD$ is given by

$V = \frac{1}{3} \cdot ($area of $\triangle ABD) \cdot ($height from $C)$.

A formal proof is very easy, and I will skip it.

Corollary

Given a tetrahedron $T$, and an edge $e_1$ of it, we can find another tetrahedron $U$ such that $\mathbf{volume}(U) > \mathbf{volume}(T)$, with an edge $f_1 > e_1$, and such that all the other edges of $U$ are equal to the corresponding edges of $T$, $\mathbf{unless}$ the edge $e_1$ stretches between sides of $T$ which are perpendicular. When we chose a bigger $f_1$, if $e_1 < 1$ we can choose $f_1 = 1$. Or, we can choose $f_1$ such that it stretches between sides which are perpendicular.

(By "stretches between two sides" I mean that the end points of the edge are the vertices on the two sides which are not common to the two sides. In the picture above, $DC_2$ stretches between the sides $ABD, AC_2B$ of $ABC_2D$.)

Lemma

Assume we have a tetrahedron $T$ with edges $e_1, \dots, e_6$, such that $e_2, \dots, e_6 \le 1$. If there is an edge $e_m < 1$ among $e_2, \dots, e_6$ then there is a tetrahedron $U$ with volume bigger than the volume of $T$, whose edges are equal to those of $T$, except for $e_m$, which is replaced by an edge of size $1$.

Proof

Case 1: If $T$ does not have any sides which are perpendicular, then the existence of $U$ follows from the corollary.

Case 2: Assume $T$ has exactly two sides which are perpendicular (like $ABC_2D$ in the picture above). If $C_2D < 1$ and it were the only edge $< 1$, then all the other edges are $= 1$ (because they were assumed to be $\le 1$). In this case $\triangle ABD, \triangle AC_2B$ are equilateral with sides $= 1$, and the planes can not be perpendicular since $C_2D < 1$. (Indeed, an easy computation shows that if we take two equilateral triangles $\triangle ABD, \triangle AC_2B$ and place them perpendicular to each other, then $CD = \frac{\sqrt{6}}{2} > 1$.) So $e_m$ must be one of the other sides. Then again, the existence of $U$ follows from the corollary.

Case 3: Assume that three sides are perpendicular.

Prob 1967 2 fig3.png

Assume the perpendicular sides are the ones meeting at $A$, i.e. each pair of the planes meeting at $A$ are perpendicular. Since at least two of the edges $BD, BC, CD \le 1$ it follows that $AB, AC, AD < 1$ (the sides of the right angle in a right triangle are less than the hypotenuse). Just for the sake of notation, assume $e_m = AB < 1$. We can apply the corollary, and find a tetrahedron $U$ with volume bigger than the volume of $T$, with edges equal to those of $T$, except that $AB$ is replaced by an edge $= 1$.

Now the problem is very easy to prove. Let $T$ be a tetrahedron with edges $e_1, \dots, e_6$, such that $e_2, \dots, e_6 \le 1$. Apply the lemma as many times as necessary (up to five times), successively replacing each edge $< 1$ by an edge $= 1$. We obtain a tetrahedron $U$ with five edges $f_2, \dots, f_6 = 1$, and one edge $f_1 = e_1$. If $f_1$ stretches between two perpendicular sides, we are done. If not, apply the corollary one more time to obtain a bigger tetrahedron in which $f_1$ is replaced by a larger edge which stretches between two perpendicular sides.

We obtain the same result as in the first solution: the largest tetrahedron is the one formed by two equilateral triangles with sides $= 1$, having one side in common, with the two planes containing the triangles perpendicular. An easy calculation shows that the edge which is $> 1$ is in fact of length $\frac{\sqrt{6}}{2}$, and the volume of this tetrahedron is $\frac{1}{8}$.

[Solution by pf02, September 2024]


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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