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1969 Canadian MO Problems/Problem 5: Difference between revisions

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== Problem ==
== Problem ==
Let <math>\displaystyle ABC</math> be a triangle with sides of length <math>\displaystyle a</math>, <math>\displaystyle b</math> and <math>\displaystyle c</math>. Let the bisector of the <math>\displaystyle \angle C</math> cut <math>\displaystyle AB</math> at <math>\displaystyle D</math>. Prove that the length of <math>\displaystyle CD</math> is <math>\displaystyle \frac{2ab\cos \frac{C}{2}}{a+b}.</math>
Let <math>ABC</math> be a triangle with sides of length <math>a</math>, <math>b</math> and <math>c</math>. Let the bisector of the <math>\angle C</math> cut <math>AB</math> at <math>D</math>. Prove that the length of <math>CD</math> is <math>\frac{2ab\cos \frac{C}{2}}{a+b}.</math>




== Solution ==
== Solution ==
Let <math>\displaystyle CD=d.</math> Note that <math>\displaystyle [\triangle ABC]=[\triangle ACD]+[\triangle BCD].</math> This can be rewritten as <math>\displaystyle \frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .</math>
Let <math>CD=d.</math> Note that <math>[\triangle ABC]=[\triangle ACD]+[\triangle BCD].</math> This can be rewritten as <math>\frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .</math>


Because <math>\displaystyle \sin C=2\sin \frac C2 \cos \frac C2,</math> the expression can be written as <math>\displaystyle2ab\cos \frac C2=d(a+b).</math> Dividing by <math>\displaystyle a+b,</math> <math>CD=\frac{2ab\cos \frac C2}{a+b},</math> as desired.
Because <math>\sin C=2\sin \frac C2 \cos \frac C2,</math> the expression can be written as <math>2ab\cos \frac C2=d(a+b).</math> Dividing by <math>a+b,</math> <math>CD=\frac{2ab\cos \frac C2}{a+b},</math> as desired.


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{{Old CanadaMO box|num-b=1|num-a=3|year=1969}}
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Revision as of 21:40, 17 November 2007

Problem

Let $ABC$ be a triangle with sides of length $a$, $b$ and $c$. Let the bisector of the $\angle C$ cut $AB$ at $D$. Prove that the length of $CD$ is $\frac{2ab\cos \frac{C}{2}}{a+b}.$


Solution

Let $CD=d.$ Note that $[\triangle ABC]=[\triangle ACD]+[\triangle BCD].$ This can be rewritten as $\frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .$

Because $\sin C=2\sin \frac C2 \cos \frac C2,$ the expression can be written as $2ab\cos \frac C2=d(a+b).$ Dividing by $a+b,$ $CD=\frac{2ab\cos \frac C2}{a+b},$ as desired.

1969 Canadian MO (Problems)
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 3


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