Difference between revisions of "1970 Canadian MO Problems/Problem 5"

Revision as of 02:03, 29 December 2007

Problem

A quadrilateral has one vertex on each side of a square of side-length 1. Show that the lengths $a$ , $b$ , $c$ and $d$ of the sides of the quadrilateral satisfy the inequalities $2\le a^2+b^2+c^2+d^2\le 4.$

Solution

Let the quadrilateral be $ABCD$ . Suppose $A$ is a distance $w$ , $1-w$ from the two nearest vertices of the square. Define $x$ , $y$ , $z$ similarly. Then the sum of the squares of the sides of the quadrilateral is $w^2 + (1-w)^2 + x^2 + (1-x)^2 + y^2 + (1-y)^2 + z^2 + (1-z)^2$ . But $w^2 + (1-w)^2 = 2(w - \frac{1}{2})^2 + \frac{1}{2}$ which is at least $\frac{1}{2}$ and at most $1$ . Similarly for the other pairs of terms, and hence proved.

1970 Canadian MO (Problems)
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •	Followed by Problem 6

@@ Line 4: / Line 4: @@
 == Solution ==
-{{solution}}
+Let the quadrilateral be <math>ABCD</math>. Suppose <math>A</math> is a distance <math>w</math>, <math>1-w</math> from the two nearest vertices of the square. Define <math>x</math>, <math>y</math>, <math>z</math> similarly. Then the sum of the squares of the sides of the quadrilateral is <math>w^2 + (1-w)^2 + x^2 + (1-x)^2 + y^2 + (1-y)^2 + z^2 + (1-z)^2</math>. But <math>w^2 + (1-w)^2 = 2(w - \frac{1}{2})^2 + \frac{1}{2}</math> which is at least <math>\frac{1}{2}</math> and at most <math>1</math>. Similarly for the other pairs of terms, and hence proved.
 {{Old CanadaMO box|num-b=4|num-a=6|year=1970}}

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Difference between revisions of "1970 Canadian MO Problems/Problem 5"

Revision as of 02:03, 29 December 2007

Problem

Solution