Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1975 USAMO Problems/Problem 2"
m (→‎Solution 3)
 
(14 intermediate revisions by 6 users not shown)
Line 4: Line 4:
 
<cmath>AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.</cmath>
 
<cmath>AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2.</cmath>
  
−
==Solution==
+
==Solutions==
  
 
===Solution 1===
 
===Solution 1===
Line 63: Line 63:
  
 
Let
 
Let
 +
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
A &= (0,0,0) \\
 
A &= (0,0,0) \\
Line 68: Line 69:
 
C &= (a,b,c) \\
 
C &= (a,b,c) \\
 
D &= (x,y,z).
 
D &= (x,y,z).
−
\end{align*}</cmath>
+
\end{align*}
 +
</cmath>
 +
 
 
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,
 +
 
<cmath>\begin{align*}
 
<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 \\
 
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*}</cmath>
+
\end{align*}
 +
</cmath>
 
which rearranges to the desired inequality.
 
which rearranges to the desired inequality.
  
−
==See also==
+
===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.
 +
 
 +
''NOTES:''
 +
 
 +
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.
 +
 
 +
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}}
 +
 
 +
==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 (Problems • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1975_USAMO_Problems/Problem_2&oldid=87218"