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Difference between revisions of "1976 IMO Problems/Problem 1"

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== Problem ==
 
== Problem ==
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{{problem}}
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In a convex quadrilateral (in the plane) with the area of <math>32 \text{ cm}^{2}</math> the sum of two opposite sides and a diagonal is <math>16 \text{ cm}</math>. Determine all the possible values that the other diagonal can have.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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Label the vertices <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> in such a way that <math>AB + BD + DC = 16</math>, and <math>\overline{BD}</math> is a diagonal.
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The area of the quadrilateral can be expressed as <math>BD \cdot ( d_1 + d_2 ) / 2</math>, where <math>d_1</math> and <math>d_2</math> are altitudes from points <math>A</math> and <math>C</math> onto <math>\overline{BD}</math>. Clearly, <math>d_1 \leq AB</math> and <math>d_2 \leq DC</math>. Hence the area is at most <math>BD \cdot ( AB + DC ) / 2 = BD(16-BD) / 2</math>.
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The quadratic function <math>f(x)=x(16-x)/2</math> has its maximum for <math>x=8</math>, and its value is <math>f(8)=32</math>.
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The area of our quadrilateral is <math>32</math>. This means that we must have <math>BD=8</math>. Also, equality must hold in both <math>d_1 \leq AB</math> and <math>d_2 \leq DC</math>. Hence both <math>\overline{AB}</math> and <math>\overline{DC}</math> must be perpendicular to <math>\overline{BD}</math>. And in any such case it is clear from the Pythagorean theorem that <math>AC = 8\sqrt 2</math>.
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Therefore the other diagonal has only one possible length: <math>8\sqrt 2</math>.
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== See also ==
 
== See also ==
 
{{IMO box|year=1976|before=First question|num-a=2}}
 
{{IMO box|year=1976|before=First question|num-a=2}}

Latest revision as of 21:05, 26 September 2009

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: $8\sqrt 2$.

See also

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