Difference between revisions of "1975 USAMO Problems/Problem 2"
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==Problem== | ==Problem== | ||
− | Let <math>A,B,C,D</math> denote four points in space and <math>AB</math> the distance between <math>A</math> and <math>B</math>, and so on. Show that < | + | |
+ | Let <math>A,B,C,D</math> denote four points in space and <math>AB</math> the distance between <math>A</math> and <math>B</math>, and so on. Show that | ||
+ | <cmath>AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.</cmath> | ||
==Solution== | ==Solution== | ||
+ | |||
+ | ===Solution 1=== | ||
+ | |||
<asy> | <asy> | ||
defaultpen(fontsize(8)); | defaultpen(fontsize(8)); | ||
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label("$a$",(A+D)/2,(1,1));label("$d$",(B+D)/2,(-1,1)); | label("$a$",(A+D)/2,(1,1));label("$d$",(B+D)/2,(-1,1)); | ||
</asy> | </asy> | ||
− | If we project points <math>A,B,C,D</math> onto the plane parallel to <math>AB</math> and <math>CD</math>, <math>AB</math> and <math>CD</math> stay the same but <math>BC, AC, AD, BD</math> all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when <math>A,B,C,D</math> are coplanar: | + | |
+ | If we project points <math>A,B,C,D</math> onto the plane parallel to <math>\overline{AB}</math> and <math>\overline{CD}</math>, <math>AB</math> and <math>CD</math> stay the same but <math>BC, AC, AD, BD</math> all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when <math>A,B,C,D</math> are coplanar: | ||
+ | |||
<asy> | <asy> | ||
size(200); | size(200); | ||
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First, we express <math>a^2+b^2</math> in terms of <math>c,d,m,\theta,\phi</math>, using the [[Law of Cosines]]: | First, we express <math>a^2+b^2</math> in terms of <math>c,d,m,\theta,\phi</math>, using the [[Law of Cosines]]: | ||
− | < | + | <cmath>\begin{align*} |
− | + | a^2+b^2 &= c^2+d^2+2m^2-2cm\cos(\theta)-2dm\cos(\phi-\theta) \ | |
− | <math>a^2+b^2</math> is a function of <math>\theta</math>, so we take the derivative with respect to <math>\theta</math> and obtain that <math>a^2+b^2</math> takes a minimum when < | + | (a^2+b^2-c^2-d^2-2m^2)^2 &= 4m^2(c^2\cos^2(\theta)+d^2\cos^2(\phi-\theta)+2cd\cos(\theta)\cos(\phi-\theta)) |
− | + | \end{align*}</cmath> | |
− | + | <math>a^2+b^2</math> is a function of <math>\theta</math>, so we take the derivative with respect to <math>\theta</math> and obtain that <math>a^2+b^2</math> takes a minimum when | |
− | &= | + | <cmath>\begin{align*} |
− | &= | + | c\sin(\theta)-d\sin(\phi-\theta) &= 0 \ |
− | \end{ | + | c^2\sin^2(\theta)+d^2\sin^2(\phi-\theta)-2cd\sin(\theta)\sin(\phi-\theta) &= 0 \ |
+ | (a^2+b^2-c^2-d^2-2m^2)^2 &= 4m^2(c^2+d^2+2cd(\cos(\theta)\cos(\phi-\theta)-\sin(\theta)\sin(\phi-\theta))) \ | ||
+ | &= 4m^2(c^2+d^2+2cd\cos{\phi})\ | ||
+ | &= 4m^2(2c^2+2d^2-n^2) | ||
+ | \end{align*}</cmath> | ||
Define <math>p=a^2+b^2</math> and <math>q=c^2+d^2</math>: | Define <math>p=a^2+b^2</math> and <math>q=c^2+d^2</math>: | ||
− | < | + | <cmath>\begin{align*} |
− | + | (p-q-2m^2)^2 &= 4m^2(2q-n^2) \ | |
− | + | p^2+q^2+4m^4-4m^2p+4m^2q-2pq &= 8m^2q-4m^2n^2 \ | |
− | + | p^2+q^2+4m^4-4m^2p-4m^2q-2pq &= -4m^2n^2 \ | |
− | + | p^2-2pq+q^2-4m^2(p+q) &= -4m^2(m^2+n^2) \ | |
− | + | \frac{(p-q)^2}{m^2} &= p+q-m^2-n^2\geq 0 \ | |
− | + | a^2+b^2+c^2+d^2 &\geq m^2+n^2 \ | |
+ | \end{align*}</cmath> | ||
− | + | ===Solution 2=== | |
− | < | + | Let |
+ | <cmath>\begin{align*} | ||
+ | A &= (0,0,0) \ | ||
+ | B &= (1,0,0) \ | ||
+ | C &= (a,b,c) \ | ||
+ | D &= (x,y,z). | ||
+ | \end{align*}</cmath> | ||
+ | It is clear that every other case can be reduced to this. | ||
+ | Then, with the distance formula and expanding, | ||
+ | <cmath>\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 \ | ||
+ | &= (x+a-1)^2 + (y+b)^2 + (z+c)^2. \ | ||
+ | &\geq 0, | ||
+ | \end{align*}</cmath> | ||
+ | which rearranges to the desired inequality. | ||
==See also== | ==See also== |
Revision as of 11:45, 2 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
It is clear that every other case can be reduced to this.
Then, with the distance formula and expanding,
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 |