1976 IMO Problems/Problem 1

Problem

In a convex quadrilateral (in the plane) with the area of $32 \text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \text{ cm}$ . Determine all the possible values that the other diagonal can have.

Solution

Label the vertices $A$ , $B$ , $C$ , and $D$ in such a way that $AB + BD + DC = 16$ , and $\overline{BD}$ is a diagonal.

The area of the quadrilateral can be expressed as $BD \cdot ( d_1 + d_2 ) / 2$ , where $d_1$ and $d_2$ are altitudes from points $A$ and $C$ onto $\overline{BD}$ . Clearly, $d_1 \leq AB$ and $d_2 \leq DC$ . Hence the area is at most $BD \cdot ( AB + DC ) / 2 = BD(16-BD) / 2$ .

The quadratic function $f(x)=x(16-x)/2$ has its maximum for $x=8$ , and its value is $f(8)=32$ .

The area of our quadrilateral is $32$ . This means that we must have $BD=8$ . Also, equality must hold in both $d_1 \leq AB$ and $d_2 \leq DC$ . Hence both $\overline{AB}$ and $\overline{DC}$ must be perpendicular to $\overline{BD}$ . And in any such case it is clear from the Pythagorean theorem that $AC = 8\sqrt 2$ .

Therefore the other diagonal has only one possible length: $8\sqrt 2$ .

1976 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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1976 IMO Problems/Problem 1

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