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

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== Problem ==
 
== Problem ==
A quadrilateral has one vertex on each side of a square of side-length 1. Show that the lengths <math>a</math>, <math>b</math>, <math>c</math> and <math>d</math> of the sides of the quadrilateral satisfy the inequalities <math>2\le a^2+b^2+c^2+d^2\le 4.</math>
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A [[quadrilateral]] has one [[vertex]] on each side of a [[square]] of side-length 1. Show that the lengths <math>a</math>, <math>b</math>, <math>c</math> and <math>d</math> of the sides of the quadrilateral satisfy the inequalities <math>2\le a^2+b^2+c^2+d^2\le 4.</math>
  
  

Latest revision as of 09:08, 20 March 2008

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