Difference between revisions of "1967 IMO Problems/Problem 2"
m |
m |
||
Line 31: | Line 31: | ||
==Solution== | ==Solution== | ||
− | Assume <math>CD > 1</math> and assume that all other sides are <math>\ le 1</math>. | + | 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> | + | 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 | ||
− | + | <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. | + | |
+ | |||
+ | |||
+ | |||
+ | |||
+ | TO BE CONTINUED. DOING A SAVE MIDWAY SO I DON'T LOOSE WORK DONE SO FAR. | ||
Revision as of 19:20, 13 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
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 |