# Geometry problem on the incentre and circumcenter of a triangle

I have the following problem:

In a triangle $ABC$ the line joining incentre and circumcentre is parallel to side $BC$. Prove that $\cos B + \cos C=1$.

Could someone help me solve it?

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This question provides simple proofs of the identity: $$\frac{r}{R}+1=\cos A+\cos B+\cos C$$

Referring to Derek's diagram, $\angle CKB = 2\angle A = 2\angle CKR$. Since, the angle made by the chord $BC$ at the point $A$ on the circumcircle of $\Delta ABC$ is half the angle made at the circumcentre $K$.

Also, $\cos A = \cos \angle CKR = \dfrac{KR}{CK}=\dfrac{r}{R}$ (since, IK is parallel to $BC$, we have $KR = r$).

So, $\cos B + \cos C = 1$

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This one has been untouched for a few days now, so here's a solution. It does the job but it feels like I beat it into submission with a stick.

I would be interested to know if anyone has a more elegant approach, as it was tougher than I was expecting.

In the above diagram $I$ is the incentre and $K$ is the circumcentre of $\Delta$ ABC.

$PS \parallel BC$ therefore

$$\frac{CP}{BS}=\frac{AC}{AB}= \frac{ \sin B}{\sin C}, \quad (1)$$

by the sine rule.

Furthermore, $\Delta$ CIP and $\Delta$ BIS are isosceles and so

$$\frac{PS}{BC}=\frac{CP+BS}{BC}= 1 - \frac{BS}{AB},$$

the last equality is implied by $PS \parallel BC .$ And so

$$CP = \left( 1 - \frac{BS}{AB} \right) BC - BS = \frac{\sin B}{\sin C} \cdot BS,$$ using $(1).$

Hence

$$BC = BS \left( 1+ \frac{\sin B}{\sin C} + \frac{BC}{AB} \right) = BS \left( 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} \right), \quad (2)$$ by the sine rule.

Let $\angle CBK=\theta$ then, again by the sine rule and noting that $BK=CK,$ we have $$\frac{\sin B}{\sin (B+C)} = \frac{AC}{BC} = \frac{\cos (C-\theta)}{\cos \theta} = \cos C + \sin C \tan \theta.$$

Hence

$$\tan \theta = \frac{ \frac{\sin B}{\sin (B+C)} - \cos C}{\sin C}.$$

Also

$$BC = 2 RK \cot \theta = 2 BS \sin B \cot \theta = \frac{2BS \sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}, \quad (3)$$

using $RK = BS \sin B$ and the expression for $\tan \theta.$

Combining $(2)$ and $(3)$ we obtain

$$1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} =\frac{2\sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}. \quad (4)$$

Now we need to show that $(4) \Rightarrow \cos B + \cos C = 1.$

ABC is a non-degenerate triangle and so $C \ne 0$ and $\cos B + \cos C \ne -1$ and so $\cos B + \cos C = 1 \Longleftrightarrow$

$$2\sin B \sin^2 C = 2 \sin B (1- \cos^2 C) + \sin C \lbrace 1 - (\cos B + \cos C)^2 \rbrace$$ $$= 2\sin B - 2\sin B \cos^2 C + \sin C (-2\cos B \cos C + \sin^2 B - \cos^2 C)$$ $$=\lbrace 2 + \cos(B-C) \rbrace \lbrace \sin B - \cos C \sin(B+C) \rbrace, \quad (5)$$

noting that $\sin(B+C)\cos(B-C) = \sin B \cos B + \sin C \cos C.$ When we expand the RHS of $(5)$ the two terms $\sin B \cos B \cos C$ and $-\sin B \cos B \cos C$ cancel, leaving precisely the RHS of the second to last line of $(5).$

Hence we have $\cos B + \cos C = 1 \Longleftrightarrow$

$$\frac{2+ \cos(B-C)}{\sin C} = \frac{2 \sin B \sin C}{\sin B - \cos C \sin (B+C)}. \quad (6)$$

Now

$$\sin (B+C) \lbrace 1+ \cos(B-C) \rbrace = \sin(B+C) + \sin(B+C) \cos(B-C)$$ $$=\sin B \cos C + \cos B\sin C + \sin B \cos B + \sin C \cos C$$ $$=(\sin B + \sin C)(\cos B + \cos C) = \sin B + \sin C \quad (7)$$

$\Longleftrightarrow \cos B + \cos C = 1,$ since $\sin B + \sin C \ne 0.$

And so the LHS of $(4)$

$$1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} = \frac{\sin B + \sin C + \sin(B+C)}{\sin C}$$ and using $(7)$ to substitute for $\sin B + \sin C$ $$=\frac{2\sin(B+C)+ \sin(B+C)\cos(B-C)}{\sin C} =\sin(B+C)\frac{2+\cos(B-C)}{\sin C}$$

and using $(6)$ to substitute for the fraction

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

And so we have shown that when $B$ and $C$ are angles of a triangle then $(4) \Rightarrow \cos B + \cos C = 1$ and the result is proved.

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