Difference between revisions of "1975 USAMO Problems/Problem 2"
(New page: ==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 <center><math>AC^2+BD^2+AD...) |
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==Solution== | ==Solution== | ||
− | {{ | + | <asy> |
+ | defaultpen(fontsize(8)); | ||
+ | pair A=(2,4), B=(0,0), C=(4,0), D=(4,2); | ||
+ | label("A",A,(0,1));label("D",D,(1,0));label("B",B,(-1,-1));label("C",C,(1,-1)); | ||
+ | axialshade(A--C--D--cycle, lightgray, A, gray, D); | ||
+ | draw(A--B--C--A--D--C);draw(B--D, linetype("8 8")); | ||
+ | label("$m$",(A+B)/2,(-1,1));label("$n$",(C+D)/2,(1,0)); | ||
+ | label("$c$",(B+C)/2,(0,-1));label("$b$",(A+C)/2,(-1,-1)); | ||
+ | label("$a$",(A+D)/2,(1,1));label("$d$",(B+D)/2,(-1,1)); | ||
+ | </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: | ||
+ | <asy> | ||
+ | size(200); | ||
+ | defaultpen(fontsize(8)); | ||
+ | pair A=(8,3), B=(4,-5), C=(10,0), D=(0,0); | ||
+ | draw(A--C--B--D--A--B);draw(C--D);draw(anglemark(A,B,D,40));draw(anglemark(C,B,A,55,60)); | ||
+ | label("A",A,(0,1));label("D",D,(-1,0));label("B",B,(0,-1));label("C",C,(1,0)); | ||
+ | label("$m$",(A+B)/2,(1,0));label("$n$",(C+D)/2,(0,1)); | ||
+ | label("$c$",(B+C)/2,(1,-1));label("$b$",(A+C)/2,(-1,-1)); | ||
+ | label("$a$",(A+D)/2,(0,1));label("$d$",(B+D)/2,(-1,-1)); | ||
+ | label("$\phi-\theta$",anglemark(A,B,D,40),(1,5));label("$\theta$",anglemark(C,B,A,55),(8,9)); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>AD=a, AC=b, BC=c, BD=d, AB=m, CD=n</math>. We wish to prove that <math>a^2+b^2+c^2+d^2\ge m^2+n^2</math>. Let us fix <math>\triangle BCD</math> and the length <math>AB</math> and let <math>A</math> vary on the circle centered at <math>B</math> with radius <math>m</math>. If we find the minimum value of <math>a^2+b^2</math>, which is the only variable quantity, and prove that it is larger than <math>m^2+n^2-c^2-d^2</math>, we will be done. | ||
+ | |||
+ | First, we express <math>a^2+b^2</math> in terms of <math>c,d,m,\theta,\phi</math>, using the [[Law of Cosines]]: | ||
+ | <center><math>a^2+b^2=c^2+d^2+2m^2-2cm\cos(\theta)-2dm\cos(\phi-\theta)</math></center> | ||
+ | <math>\implies (a^2+b^2-c^2-d^2-2m^2)^2</math> <math>=4m^2(c^2\cos^2(\theta)+d^2\cos^2(\phi-\theta)+2cd\cos(\theta)\cos(\phi-\theta))</math>. | ||
+ | <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 <center><math>c\sin(\theta)-d\sin(\phi-\theta)=0</math></center> | ||
+ | <math>\implies c^2\sin^2(\theta)+d^2\sin^2(\phi-\theta)-2cd\sin(\theta)\sin(\phi-\theta)=0</math>. | ||
+ | <math>\begin{eqnarray*}\implies(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{eqnarray*}</math> | ||
+ | |||
+ | Define <math>p=a^2+b^2</math> and <math>q=c^2+d^2</math>: | ||
+ | |||
+ | <math>(p-q-2m^2)^2=4m^2(2q-n^2)</math> | ||
+ | |||
+ | <math>\implies p^2+q^2+4m^4-4m^2p+4m^2q-2pq=8m^2q-4m^2n^2</math> | ||
+ | |||
+ | <math>\implies p^2+q^2+4m^4-4m^2p-4m^2q-2pq=-4m^2n^2</math> | ||
+ | |||
+ | <math>\implies p^2-2pq+q^2-4m^2(p+q)=-4m^2(m^2+n^2)</math> | ||
+ | |||
+ | <math>\implies \frac{(p-q)^2}{m^2}=p+q-m^2-n^2\ge 0</math> | ||
+ | |||
+ | <math>\implies a^2+b^2+c^2+d^2\ge m^2+n^2</math>. | ||
==See also== | ==See also== |
Revision as of 21:35, 3 January 2009
Problem
Let denote four points in space and the distance between and , and so on. Show that
Solution
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 :
.
See also
1975 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |