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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 $(1).$


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

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


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


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

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 $\cos B + \cos C = 1 \Longleftrightarrow$

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


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

$\Longleftrightarrow \cos B + \cos C = 1,$ 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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Nice. I had been thinking on and off about this. – Andrés E. Caicedo Feb 15 '11 at 22:49
@Andres: Thanks. When I discovered this solution I stopped searching for alternatives, but I'd be interested in seeing others. The problem is surprisingly intractable given its innocent appearance. I'm interested in knowing the origin of the problem but unfortunately the OP seems to have disappeared :-( – Derek Jennings Feb 17 '11 at 15:52

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