Difference between revisions of "1967 IMO Problems/Problem 2"
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The volume of the tetrahedron is | The volume of the tetrahedron is | ||
− | <math>V = \frac{1}{3} \cdot (</math>area of <math>\triangle ABD) \cdot</math> height from <math>C = | + | <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 | \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) = | \left( \frac{1}{6} \cdot x \cdot \sqrt{1 - \frac{x^2}{4}} \cdot \sqrt{1 - \frac{x^2}{4}} \right) = | ||
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These statements are intuitively clear, since the volume <math>V</math> of the | These statements are intuitively clear, since the volume <math>V</math> of the | ||
tetrahedron <math>ABCD</math> is given by | tetrahedron <math>ABCD</math> is given by | ||
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+ | <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. | ||
+ | |||
+ | The conclusion is that given a tetrahedron <math>T</math>, and an edge <math>e_1</math> of | ||
+ | it, we ca find another tetrahedron <math>U</math> such that <math>vol(U) > vol(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>, unless the edge <math>e_1</math> | ||
+ | stretches between sides of <math>T</math> which are perpendicular. | ||
Revision as of 01:00, 14 September 2024
Prove that if one and only one edge of a tetrahedron is greater than , then its volume is .
Contents
[hide]Solution
Assume and let . Let be the feet of perpendicular from to and and from to , respectively.
Suppose . We have that , . We also have . So the volume of the tetrahedron is .
We want to prove that this value is at most , which is equivalent to . This is true because .
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 is not really necessary. The statement is true even if all sides are . What we need is that no more than one side is .
The upper limit of for the volume of the tetrahedron is actually reached. This will become clear from both solutions.
Solution
Assume and assume that all other sides are . Let . Let be the feet of perpendiculars from to , from to the plane , and from to , respectively.
At least one of the segments has to be . Suppose . (If were bigger that the argument would be the same.) We have that . By the same argument in we have . Since plane , we have , so .
The volume of the tetrahedron is
area of (height from .
We need to prove that . Some simple computations show that this is the same as . This is true because , and on this interval.
Note: is achieved when and all inequalities are equalities. This is the case when all sides except are , are midpoints of and (in which case the planes are perpendicular). In this case, , as can be seen from an easy computation.
Solution 2
We begin with a simple observation. Let be a tetrahedron, and consider the transformations which rotate around while keeping fixed. We get a set of tetrahedrons, two of which, and are shown in the picture below. The lengths of all sides except are constant through this transformation.
1. Assume that the angles between the planes and , and and are both acute. If the perpendicular from to the plane is larger that the perpendicular from to the plane then the volume of is larger than the volume of .
2. Furthermore, the tetrahedron obtained when the position of is such that the planes and are perpendicular has the maximum volume of all tetrahedrons obtained from rotating around .
These statements are intuitively clear, since the volume of the tetrahedron is given by
area of height from . A formal proof is very easy, and I will skip it.
The conclusion is that given a tetrahedron , and an edge of it, we ca find another tetrahedron such that , with an edge , and such that all the other edges of are equal to the corresponding edges of , unless the edge stretches between sides of which are perpendicular.
(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 |