Hello arze Originally Posted by

**arze** ABC is a triangle with no obtuse angles. The perpendicular AD from A to BC meets the circumcircle of the triangle at E. If the incentre of the triangle is equidistant from A and E, prove that $\displaystyle \cos B+\cos C=1$.

I have no idea how to begin.

thanks

This is *much, much* harder than the other trig questions you've posted. I wonder where it came from.

I'm afraid this Originally Posted by

**sa-ri-ga-ma** If I is the incenter, and AI = IE, then I must lie on the perpendicular bisector of AE and it must pass through circumcenter of ABC.

Secondly I must be equidistant from all the three sides of the triangle ABC.

These two conditions are satisfied only when ABC is **equilateral** triangle. In that case B = C = 60 degrees and cosB + cosC = 1.

isn't true. The triangle ABC needn't be equilateral.

I have a proof - see below - but it's very complicated for such an apparently simple result. Perhaps someone may be able to come up with something more straightforward?

Here goes.

Look at the attached diagram. I is the incentre.

I haven't drawn the circumcircle - the only property of this that I'm using is that:$\displaystyle \angle AEC = \angle ABC$ (angles in same segment) Call this equation (1)

Then we need to express $\displaystyle AE$ in two ways:1. In terms of the sides and angles of any triangle $\displaystyle ABC$.

2. Using the fact that $\displaystyle \triangle AEI$ is isosceles, with $\displaystyle AI = IE$.

So first:$\displaystyle AD= b \sin C$ ...(2)

$\displaystyle DC = b \cos C$ ...(3)

$\displaystyle DE = DC \cot \angle DEC = DC \cot B$, from (1)$\displaystyle = \frac{b\cos C \cos B}{\sin B}$, from (3)

$\displaystyle \Rightarrow AE = AD +DE$$\displaystyle = b\sin C + \frac{b\cos C \cos B}{\sin B}$, from (2)

$\displaystyle = \frac{b\cos(B-C)}{\sin B}$, using $\displaystyle \cos (B-C) = \cos B \cos C +\sin B \sin C$. Call this equation (4)

Then, expressing $\displaystyle AE$ using the fact that $\displaystyle \triangle AEI$ is isosceles:

The radius of the incircle, $\displaystyle r$, is given by:$\displaystyle r = AK =\frac{2\Delta}{a+b+c}$, where $\displaystyle \Delta$ is the area of the triangle ABC

Now $\displaystyle AI$ bisects $\displaystyle \angle BAC$. So $\displaystyle \angle IAK = \tfrac12A$$\displaystyle \Rightarrow AI = \frac{AK}{\sin\tfrac12A}$$\displaystyle =\frac{2\Delta}{(a+b+c)\sin\tfrac12A}$ ...(5)

Also, from $\displaystyle \triangle ABD,\; \angle BAD = 90^o-B$$\displaystyle \Rightarrow \angle JAI = \tfrac12A - (90^o-B) $$\displaystyle =\tfrac12A+B-90^o$

So $\displaystyle AE = 2AI\cos\angle JAI$$\displaystyle =2AI \cos(\tfrac12A+B-90^o)$

$\displaystyle =2AI\sin(\tfrac12A+B)$

$\displaystyle = \frac{4\Delta\sin(\tfrac12A+B)}{(a+b+c)\sin\tfrac1 2A}$, from (5)

So, from (4), equating these two expressions for $\displaystyle AE$, using (4):$\displaystyle \frac{b\cos(B-C)}{\sin B}=\frac{4\Delta\sin(\tfrac12A+B)}{(a+b+c)\sin\tfr ac12A}$

We now use $\displaystyle \Delta = \tfrac12bc\sin A$, and then use the Sine Rule to write $\displaystyle a$ and $\displaystyle b$ in terms of $\displaystyle c$:$\displaystyle \Rightarrow\frac{b\cos(B-C)}{\sin B}=\frac{2bc\sin A\sin(\tfrac12A+B)}{(\frac{c\sin A}{\sin C}+\frac{c\sin B}{\sin C}+c)\sin\tfrac12A}$$\displaystyle =\frac{2b\sin A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)\sin\tfrac12A}$

$\displaystyle \Rightarrow\frac{\cos(B-C)}{\sin B}=\frac{4\sin\tfrac12 A\cos\tfrac12A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)\sin\tfrac12A}$$\displaystyle =\frac{4\cos\tfrac12A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)}$

All that remains now - and there's still quite a bit to do - is to eliminate $\displaystyle A$ using $\displaystyle A = 180^o-(B+C)$, and the result should follow. So, noting that:$\displaystyle \cos\tfrac12A = \cos(90-\tfrac12[B+C])$$\displaystyle =\sin\tfrac12(B+C)$

and$\displaystyle \sin(\tfrac12A+B) = \sin(90-\tfrac[B+C]+B)$$\displaystyle =\cos\tfrac12(B-C)$

and$\displaystyle \sin A = \sin (B+C)$

we get:$\displaystyle \frac{\cos(B-C)}{\sin B}=\frac{4\sin\tfrac12(B+C)\sin C\cos\tfrac12(B-C)}{\sin(B+C)+\sin B + \sin C}$$\displaystyle =\frac{4\sin\tfrac12(B+C)\sin C\cos\tfrac12(B-C)}{2\sin\tfrac12(B+C)\cos\tfrac12(B+C)+2\sin\tfra c12(B + C)\cos\tfrac12(B-C)}$

$\displaystyle =\frac{2\sin C\cos\tfrac12(B-C)}{\cos\tfrac12(B+C)+\cos\tfrac12(B-C)}$

$\displaystyle =\frac{4\sin\tfrac12 C\cos\tfrac12C\cos\tfrac12(B-C)}{2\cos\tfrac12B\cos\tfrac12C}$

$\displaystyle \Rightarrow \frac{2\cos^2\tfrac12(B-C)-1}{2\sin\tfrac12B\cos\tfrac12B}=\frac{2\sin\tfrac1 2C\cos\tfrac12(B-C)}{\cos\tfrac12B}$

$\displaystyle \Rightarrow 2\cos^2\tfrac12(B-C)-1=4\sin\tfrac12B\sin\tfrac12C\cos\tfrac12(B-C)$

$\displaystyle \Rightarrow 2\cos\tfrac12(B-C)\Big(\cos\tfrac12(B-C)-2\sin\tfrac12B\sin\tfrac12C\Big)=1$

$\displaystyle \Rightarrow2\cos\tfrac12(B-C)(\cos\tfrac12B\cos\tfrac12C+\sin\tfrac12B\sin\tf rac12C-2\sin\tfrac12B\sin\tfrac12C)=1$

$\displaystyle \Rightarrow2\cos\tfrac12(B-C)(\cos\tfrac12B\cos\tfrac12C-\sin\tfrac12B\sin\tfrac12C)=1$

$\displaystyle \Rightarrow2\cos\tfrac12(B-C)\cos\tfrac12(B+C)=1$

$\displaystyle \Rightarrow \cos B + \cos C = 1$

Phew!

You would think that there was an easier way, but I can't see it!

Grandad