Difference between revisions of "1975 USAMO Problems/Problem 2"

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.

1975 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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