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> | + | 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> |
== 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. | ||
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Latest revision as of 08:08, 20 March 2008
Problem
A quadrilateral has one vertex on each side of a square of side-length 1. Show that the lengths 
, 
, 
and 
of the sides of the quadrilateral satisfy the inequalities 
Solution
Let the quadrilateral be 
. Suppose 
is a distance 
, 
from the two nearest vertices of the square. Define 
, 
, 
similarly. Then the sum of the squares of the sides of the quadrilateral is 
. But 
which is at least 
and at most 
. 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 |
