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>
 
  
==Solution==
+
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>
 +
 
 +
==Solutions==
 +
 
 +
===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]]:
<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) \
 +
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.
 +
 
 +
===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 <math> x </math>, <math>y </math>, and <math> z </math> dimension to get the desired result. Let <math> x_a</math>, <math>x_b</math>, <math>x_c</math>, and <math>x_d</math> be the positions of <math> A </math>, <math>B</math>, <math>C</math>, and <math>D</math> respectively. Then we must show that,
 +
 
 +
<cmath>\begin{align*}
 +
(x_a - x_c)^2 + (x_b-x_d)^2 + (x_a - x_d)^2 + (x_b - x_c)^2 &\geq (x_a - x_b)^2 + (x_c-x_d)^2 \
 +
x_a^2 + x_b^2 + x_c^2 + x_d^2 &\geq 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 &\geq 2(x_a +x_b)(x_c + x_d)\
 +
(x_a + x_b - x_c - x_d)^2 &\geq 0.
 +
\end{align*}
 +
</cmath>
 +
So we are done.
 +
 
 +
===Solution 4 (Vector bash)===
 +
Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> correspond to the position vectors of points A, B, C, and D, respectively, with respect to an arbitrary origin O. Let us also for simplicity define <math>a^2 = a \cdot a = ||a||^2</math>, where <math>||a||</math> is the magnitude of vector <math>a</math>. Because squares are non-negative, <math>a^2</math> is non-negative for all vectors <math>a</math>. Thus,
 +
<cmath>(a + b - c - d)^2 \ge 0</cmath>
 +
Because dot product is linear, we expand to obtain
 +
<cmath>a^2 + b^2 + c^2 + d^2 + 2a \cdot b + 2 c \cdot d - 2 a \cdot c - 2 a \cdot d - 2 b \cdot c - 2 c \cdot d \ge 0,</cmath>
 +
from which we add <math>a^2 + b^2 + c^2 + d^2</math> to both sides, rearrange, and complete the square to get
 +
<cmath>(a-c)^2 + (a-d)^2 + (b-c)^2 + (b-d)^2 \ge (a-b)^2 + (c-d)^2.</cmath>
 +
As <math>(a-b)^2 = ||a-b||^2 = ||AB||^2 = AB^2</math> and likewise for the others,
 +
<cmath>AC^2 + AD^2 + BC^2 + BD^2 \ge AB^2 + CD^2,</cmath>
 +
which is what we wanted to prove.
  
<math>\implies p^2+q^2+4m^4-4m^2p+4m^2q-2pq=8m^2q-4m^2n^2</math>
+
''NOTES:''
  
<math>\implies p^2+q^2+4m^4-4m^2p-4m^2q-2pq=-4m^2n^2</math>
+
1. Equality holds when the vector equality <math>a + b = c + d</math> holds, which occurs when A, B, C, and D are the vertices of a (planar) parallelogram, in that order.
  
<math>\implies p^2-2pq+q^2-4m^2(p+q)=-4m^2(m^2+n^2)</math>
+
2. The algebra employed here is almost identical to the algebra used in Solution 3, which means that Solution 3 is just a simplification of this solution.
  
<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>.
+
{{alternate solutions}}
  
==See also==
+
==See Also==
  
 
{{USAMO box|year=1975|num-b=1|num-a=3}}
 
{{USAMO box|year=1975|num-b=1|num-a=3}}
 +
{{MAA Notice}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]
 +
[[Category:3D Geometry Problems]]

Latest revision as of 22:07, 26 August 2017

Problem

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

Solutions

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.

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 $x$, $y$, and $z$ dimension to get the desired result. Let $x_a$, $x_b$, $x_c$, and $x_d$ be the positions of $A$, $B$, $C$, and $D$ respectively. Then we must show that,

\begin{align*} (x_a - x_c)^2 + (x_b-x_d)^2 + (x_a - x_d)^2 + (x_b - x_c)^2 &\geq (x_a - x_b)^2 + (x_c-x_d)^2 \\ x_a^2 + x_b^2 + x_c^2 + x_d^2 &\geq 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 &\geq 2(x_a +x_b)(x_c + x_d)\\ (x_a + x_b - x_c - x_d)^2 &\geq 0. \end{align*} So we are done.

Solution 4 (Vector bash)

Let $a$, $b$, $c$, $d$ correspond to the position vectors of points A, B, C, and D, respectively, with respect to an arbitrary origin O. Let us also for simplicity define $a^2 = a \cdot a = ||a||^2$, where $||a||$ is the magnitude of vector $a$. Because squares are non-negative, $a^2$ is non-negative for all vectors $a$. Thus, \[(a + b - c - d)^2 \ge 0\] Because dot product is linear, we expand to obtain \[a^2 + b^2 + c^2 + d^2 + 2a \cdot b + 2 c \cdot d - 2 a \cdot c - 2 a \cdot d - 2 b \cdot c - 2 c \cdot d \ge 0,\] from which we add $a^2 + b^2 + c^2 + d^2$ to both sides, rearrange, and complete the square to get \[(a-c)^2 + (a-d)^2 + (b-c)^2 + (b-d)^2 \ge (a-b)^2 + (c-d)^2.\] As $(a-b)^2 = ||a-b||^2 = ||AB||^2 = AB^2$ and likewise for the others, \[AC^2 + AD^2 + BC^2 + BD^2 \ge AB^2 + CD^2,\] which is what we wanted to prove.

NOTES:

1. Equality holds when the vector equality $a + b = c + d$ holds, which occurs when A, B, C, and D are the vertices of a (planar) parallelogram, in that order.

2. The algebra employed here is almost identical to the algebra used in Solution 3, which means that Solution 3 is just a simplification of this solution.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1975 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png