1975 USAMO Problems/Problem 2
Problem
Let denote four points in space and the distance between and , and so on. Show that
Solution
Solution 1
If we project points onto the plane parallel to and , and stay the same but all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when are coplanar:
Let . We wish to prove that . Let us fix and the length and let vary on the circle centered at with radius . If we find the minimum value of , which is the only variable quantity, and prove that it is larger than , we will be done.
First, we express in terms of , using the Law of Cosines: is a function of , so we take the derivative with respect to and obtain that takes a minimum when
Define and :
Solution 2
Let
A &= (0,0,0) \ B &= (1,0,0) \ C &= (a,b,c) \ D &= (x,y,z).
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)It is clear that every other case can be reduced to this. Then, with the distance formula and expanding,
AC^2 + BD^2 + AD^2 + BC^2 - AB^2 - CD^2 &= x^2-2x+1+y^2+z^2+a^2-2a+b^2+c^2+2ax+2by+2cz \ &= (x+a-1)^2 + (y+b)^2 + (z+c)^2. \ &\geq 0,
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)which rearranges to the desired inequality.
Solution 3
Because the distances are all squared, we must only prove the result in one dimension, and then we can just add up the three individual inequalities for the , , and dimension to get the desired result. Let , , , and be the positions of , , , and respectively. Then we must show that,
(x_a - x_c)^2 + (x_b-x_d)^2 + (x_a - x_d)^2 + (x_b - x_c)^2 \ge (x_a - x_b)^2 + (x_c-x_d)^2 \ x_a^2 + x_b^2 + x_c^2 + x_d^2 \ge 2x_a x_c + 2x_b x_d + 2x_a x_d + 2x_b x_c - 2x_a x_b - 2x_c x_d \ (x_a + x_b)^2 + (x_c + x_d)^2 \ge 2(x_a +x_b)(x_c + x_d)\ (x_a + x_b - x_c - x_d)^2. \ge 0
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)So we are done.
See also
1975 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |