Difference between revisions of "1975 USAMO Problems/Problem 2"
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Let | Let | ||
− | < | + | |
+ | <center><math>\begin{align*} | ||
A &= (0,0,0) \ | A &= (0,0,0) \ | ||
B &= (1,0,0) \ | B &= (1,0,0) \ | ||
C &= (a,b,c) \ | C &= (a,b,c) \ | ||
D &= (x,y,z). | D &= (x,y,z). | ||
− | \end{align*}</ | + | \end{align*}</math></center> |
+ | |||
It is clear that every other case can be reduced to this. | It is clear that every other case can be reduced to this. | ||
Then, with the distance formula and expanding, | Then, with the distance formula and expanding, | ||
− | < | + | |
+ | <center><math>\begin{align*} | ||
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 \ | 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. \ | &= (x+a-1)^2 + (y+b)^2 + (z+c)^2. \ | ||
&\geq 0, | &\geq 0, | ||
− | \end{align*}</ | + | \end{align*}</math></center> |
+ | |||
which rearranges to the desired inequality. | which rearranges to the desired inequality. | ||
Revision as of 21:40, 21 April 2010
Contents
[hide]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.
See also
1975 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |