Difference between revisions of "1969 Canadian MO Problems/Problem 5"
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== Problem == | == Problem == | ||
− | Let <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> | + | 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> | + | 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. |
− | - | + | {{Old CanadaMO box|num-b=4|num-a=6|year=1969}} |
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Latest revision as of 12:35, 10 September 2008
Problem
Let be a triangle with sides of length
,
and
. Let the bisector of the
cut
at
. Prove that the length of
is
Solution
Let Note that
This can be rewritten as
Because the expression can be written as
Dividing by
as desired.
1969 Canadian MO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 6 |