Difference between revisions of "1969 Canadian MO Problems/Problem 5"
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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. | 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= | + | {{Old CanadaMO box|num-b=4|num-a=6|year=1969}} |
Latest revision as of 11:35, 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 ![$[\triangle ABC]=[\triangle ACD]+[\triangle BCD].$](http://latex.artofproblemsolving.com/4/9/f/49f64309b971755578af1787db24ed38fae8e0af.png)
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 |
