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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 <center><math>AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2</math>.</center>
+
 
 +
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));
Line 13: Line 18:
 
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]]:
−
<center><math>a^2+b^2=c^2+d^2+2m^2-2cm\cos(\theta)-2dm\cos(\phi-\theta)</math></center>
+
<cmath>\begin{align*}
−
<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>.
+
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 <center><math>c\sin(\theta)-d\sin(\phi-\theta)=0</math></center>
+
(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))
−
<math>\implies c^2\sin^2(\theta)+d^2\sin^2(\phi-\theta)-2cd\sin(\theta)\sin(\phi-\theta)=0</math>.
+
\end{align*}</cmath>
−
<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)))\\
+
<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  
−
&=&4m^2(c^2+d^2+2cd\cos{\phi})\\
+
<cmath>\begin{align*}
−
&=&4m^2(2c^2+2d^2-n^2)
+
c\sin(\theta)-d\sin(\phi-\theta) &= 0 \\
−
\end{eqnarray*}</math>
+
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>:
  
−
<math>(p-q-2m^2)^2=4m^2(2q-n^2)</math>
+
<cmath>\begin{align*}
−
 
+
(p-q-2m^2)^2 &= 4m^2(2q-n^2) \\
−
<math>\implies p^2+q^2+4m^4-4m^2p+4m^2q-2pq=8m^2q-4m^2n^2</math>
+
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 \\
−
<math>\implies p^2+q^2+4m^4-4m^2p-4m^2q-2pq=-4m^2n^2</math>
+
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 \\
−
<math>\implies p^2-2pq+q^2-4m^2(p+q)=-4m^2(m^2+n^2)</math>
+
a^2+b^2+c^2+d^2 &\geq m^2+n^2 \\
 +
\end{align*}</cmath>
  
−
<math>\implies \frac{(p-q)^2}{m^2}=p+q-m^2-n^2\ge 0</math>
+
===Solution 2===
  
−
<math>\implies a^2+b^2+c^2+d^2\ge m^2+n^2</math>.
+
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 12:45, 2 April 2010

Problem

Let $A,B,C,D$ denote four points in space and $AB$ the distance between $A$ and $B$, and so on. Show that \[AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.\]

Solution

Solution 1

[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 $A,B,C,D$ onto the plane parallel to $\overline{AB}$ and $\overline{CD}$, $AB$ and $CD$ stay the same but $BC, AC, AD, BD$ all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when $A,B,C,D$ 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 $AD=a, AC=b, BC=c, BD=d, AB=m, CD=n$. We wish to prove that $a^2+b^2+c^2+d^2\ge m^2+n^2$. Let us fix $\triangle BCD$ and the length $AB$ and let $A$ vary on the circle centered at $B$ with radius $m$. If we find the minimum value of $a^2+b^2$, which is the only variable quantity, and prove that it is larger than $m^2+n^2-c^2-d^2$, we will be done.

First, we express $a^2+b^2$ in terms of $c,d,m,\theta,\phi$, using the Law of Cosines: \begin{align*} a^2+b^2 &= c^2+d^2+2m^2-2cm\cos(\theta)-2dm\cos(\phi-\theta) \\ (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*} $a^2+b^2$ is a function of $\theta$, so we take the derivative with respect to $\theta$ and obtain that $a^2+b^2$ takes a minimum when \begin{align*} c\sin(\theta)-d\sin(\phi-\theta) &= 0 \\ 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*}

Define $p=a^2+b^2$ and $q=c^2+d^2$:

\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*}

Solution 2

Let \begin{align*} A &= (0,0,0) \\ B &= (1,0,0) \\ C &= (a,b,c) \\ D &= (x,y,z). \end{align*} It is clear that every other case can be reduced to this. Then, with the distance formula and expanding, \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*} 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
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