Difference between revisions of "1976 IMO Problems/Problem 1"
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− | + | == Problem == | |
+ | 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. | ||
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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>. | ||
+ | |||
+ | == See also == | ||
+ | {{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 the sum of two opposite sides and a diagonal is
. Determine all the possible values that the other diagonal can have.
Solution
Label the vertices ,
,
, and
in such a way that
, and
is a diagonal.
The area of the quadrilateral can be expressed as , where
and
are altitudes from points
and
onto
. Clearly,
and
. Hence the area is at most
.
The quadratic function has its maximum for
, and its value is
.
The area of our quadrilateral is . This means that we must have
. Also, equality must hold in both
and
. Hence both
and
must be perpendicular to
. And in any such case it is clear from the Pythagorean theorem that
.
Therefore the other diagonal has only one possible length: .
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 |