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. |
{{Old CanadaMO box|num-b=4|num-a=6|year=1970}} | {{Old CanadaMO box|num-b=4|num-a=6|year=1970}} |
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 ,
,
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 |