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and 
Prove that
Let and 
Prove that
Let and 
Prove that
Let and 
Prove that
is the incentre and point 
is defined such that 
and 
is the perpendicular bisector of 
unit apart. What is the area of the triangle 
?
.
and Shoelace gives ![$[ABC]=\frac52$](http://latex.artofproblemsolving.com/4/7/8/478cbaa2d0ce2e0225b01fc3dc655d3807abb775.png)
.
, 
, and 
. By the distance formula, 
and 
. It is well known that 
We can simplify this to ![$$[ABC] = \frac {5} {2} \sqrt{10} \sin {C}$$](http://latex.artofproblemsolving.com/d/0/c/d0cf9c28280e77a440a619e9a798ee4b1ab1c3c4.png)
The slope of 
is 
and the slope of 
is 
. Through some easy calculations, we find that 
and our final answer is 
where 
is the radius of the incircle. Since 
the semiperimeter of this triangle is 
We now find the equations of the angle bisectors of 
and 
since they intersect at the incenter.
is 
and the slope of 
is 
Therefore, the slope of their vector sum is 
We now solve 
to solve for the slope of the angle bisector of 
Let 
We have 
which rearranges to 
. Solving this quadratic, we get 
It is clear that the desired slope is negative, so we have the slope is 
Since the line passes through 
we have the equation is 
The slopes of 
and are 
and respectively, so the slope of their vector sum is 
Then, we solve 
as above. Solving, we get 
It is clear the slope is negative, so we have the slope is 
Letting the equation of the line being 
we get 
so 
.
and 
Solving, we get 
and 
We then have 
's equation is 
or 
Using distance from point to line, we have 
Then 
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![$\int_{0}^{1}\left(\left(-\frac{3}{4}x+3\right)-\left(-2x+3\right)\right)dx+\int_{1}^{4}\left(\left(-\frac{3}{4}x+3\right)-\left(-\frac{1}{3}x+\frac{4}{3}\right)\right)dx=\left.\frac58x^2\right]^1_0+\left.-\frac{5}{24}x^{2}+\frac{5}{3}x\right]^4_1=\frac52$](http://latex.artofproblemsolving.com/1/b/b/1bbe06f725db36d6da23e576f2e7f6ef8851c236.png)
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