# Triangle proof

• February 27th 2010, 06:39 AM
arze
Triangle proof
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 $\cos B+\cos C=1$.
I have no idea how to begin.
thanks
• February 27th 2010, 11:13 PM
sa-ri-ga-ma
Quote:

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 $\cos B+\cos C=1$.
I have no idea how to begin.
thanks

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.
• February 28th 2010, 10:26 AM
Hello arze
Quote:

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 $\cos B+\cos C=1$.
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
Quote:

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:
$\angle AEC = \angle ABC$ (angles in same segment) Call this equation (1)
Then we need to express $AE$ in two ways:
1. In terms of the sides and angles of any triangle $ABC$.

2. Using the fact that $\triangle AEI$ is isosceles, with $AI = IE$.
So first:
$AD= b \sin C$ ...(2)

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

$DE = DC \cot \angle DEC = DC \cot B$, from
(1)
$= \frac{b\cos C \cos B}{\sin B}$, from (3)
$\Rightarrow AE = AD +DE$
$= b\sin C + \frac{b\cos C \cos B}{\sin B}$, from (2)

$= \frac{b\cos(B-C)}{\sin B}$, using $\cos (B-C) = \cos B \cos C +\sin B \sin C$. Call this equation (4)
Then, expressing $AE$ using the fact that $\triangle AEI$ is isosceles:

The radius of the incircle, $r$, is given by:
$r = AK =\frac{2\Delta}{a+b+c}$, where $\Delta$ is the area of the triangle ABC
Now $AI$ bisects $\angle BAC$. So $\angle IAK = \tfrac12A$
$\Rightarrow AI = \frac{AK}{\sin\tfrac12A}$
$=\frac{2\Delta}{(a+b+c)\sin\tfrac12A}$ ...(5)
Also, from $\triangle ABD,\; \angle BAD = 90^o-B$
$\Rightarrow \angle JAI = \tfrac12A - (90^o-B)$
$=\tfrac12A+B-90^o$
So $AE = 2AI\cos\angle JAI$
$=2AI \cos(\tfrac12A+B-90^o)$

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

$= \frac{4\Delta\sin(\tfrac12A+B)}{(a+b+c)\sin\tfrac1 2A}$, from
(5)
So, from (4), equating these two expressions for $AE$, using (4):
$\frac{b\cos(B-C)}{\sin B}=\frac{4\Delta\sin(\tfrac12A+B)}{(a+b+c)\sin\tfr ac12A}$
We now use $\Delta = \tfrac12bc\sin A$, and then use the Sine Rule to write $a$ and $b$ in terms of $c$:
$\Rightarrow\frac{b\cos(B-C)}{\sin B}=\frac{2bc\sin A\sin(\tfrac12A+B)}{(\frac{c\sin A}{\sin C}+\frac{c\sin B}{\sin C}+c)\sin\tfrac12A}$
$=\frac{2b\sin A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)\sin\tfrac12A}$
$\Rightarrow\frac{\cos(B-C)}{\sin B}=\frac{4\sin\tfrac12 A\cos\tfrac12A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)\sin\tfrac12A}$
$=\frac{4\cos\tfrac12A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)}$
All that remains now - and there's still quite a bit to do - is to eliminate $A$ using $A = 180^o-(B+C)$, and the result should follow. So, noting that:
$\cos\tfrac12A = \cos(90-\tfrac12[B+C])$
$=\sin\tfrac12(B+C)$
and
$\sin(\tfrac12A+B) = \sin(90-\tfrac[B+C]+B)$
$=\cos\tfrac12(B-C)$
and
$\sin A = \sin (B+C)$
we get:
$\frac{\cos(B-C)}{\sin B}=\frac{4\sin\tfrac12(B+C)\sin C\cos\tfrac12(B-C)}{\sin(B+C)+\sin B + \sin C}$
$=\frac{4\sin\tfrac12(B+C)\sin C\cos\tfrac12(B-C)}{2\sin\tfrac12(B+C)\cos\tfrac12(B+C)+2\sin\tfra c12(B + C)\cos\tfrac12(B-C)}$

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

$=\frac{4\sin\tfrac12 C\cos\tfrac12C\cos\tfrac12(B-C)}{2\cos\tfrac12B\cos\tfrac12C}$
$\Rightarrow \frac{2\cos^2\tfrac12(B-C)-1}{2\sin\tfrac12B\cos\tfrac12B}=\frac{2\sin\tfrac1 2C\cos\tfrac12(B-C)}{\cos\tfrac12B}$

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

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

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

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

$\Rightarrow \cos B + \cos C = 1$
Phew!

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

• February 28th 2010, 12:26 PM
Opalg
I also found this a tough problem. I have a cumbersome solution that I held back from posting in the hope that someone would find a neater way of doing it. No offence to Grandad, but his solution is about as cumbersome as mine. Just for the record, here's an outline of how I did it.

The incentre I is on the perpendicular bisector of AE. So is the circumcentre O. So the line IO is parallel to BC (see the attached modified version of Grandad's diagram). My proof is based on the fact that the triangles BOC and BIC have the same base BC and the same vertical height, and therefore the same area.

The area of BOC is $\tfrac12aR\cos A$, where R is the circumradius and a is the length BC. But $\frac a{\sin A} = 2R$. Also, $\sin A = \sin(B+C)$ and $\cos A = -\cos(B+C)$. So the area of triangle BOC is $-\tfrac14a^2\frac{\cos(B+C)}{\sin(B+C)}$.

The area of BIC is $\tfrac12IB.IC\sin(\angle BIC)$. By the sine rule, $\frac{BI}{\sin\frac12C} = \frac{CI}{\sin\frac12B} = \frac{a}{\sin\frac12(B+C)}$. So the area of triangle BIC is $\tfrac12a^2\frac{\sin\frac12B\sin\frac12C}{\sin\fr ac12(B+C)}$.

Equate those two areas to see that $\frac{-\cos(B+C)}{\sin(B+C)} = \frac{2\sin\frac12B\sin\frac12C}{\sin\frac12(B+C)}$. On the left side, the numerator is $\sin B\sin C - \cos B\cos C$ and the denominator is $2\sin\tfrac12(B+C)\cos\tfrac12(B+C) = 2\sin\tfrac12(B+C)\bigl(\cos\tfrac12 B\cos\tfrac12 C - \sin\tfrac12 B\sin \tfrac12C\bigr)$. Make those substitutions and cross-multiply (noticing that $\sin\tfrac12(B+C)$ cancels in the two denominators):

\begin{aligned}\sin B\sin C - \cos B\cos C &= 4\sin\tfrac12B\sin\tfrac12C\bigl(\cos\tfrac12 B\cos\tfrac12 C - \sin\tfrac12 B\sin \tfrac12C\bigr)\\ &= \sin B\sin C -4\sin^2\tfrac12 B\sin^2 \tfrac12C.\end{aligned}

Now use the fact that $2\sin^2\theta = 1-\cos2\theta$ to write this as $\cos B\cos C = (1-\cos B)(1-\cos C)$, which simplifies to $\cos B + \cos C = 1$.

Phew! (Whew)
• February 28th 2010, 02:51 PM
arze
Thanks everyone! The proofs were really long.
Grandad, I got this from an A-level pure math textbook.