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Let $ABC$ be an acute triangle with circumcenter $O$ and incenter $I$. Points $E$, $M$ lie on $AC$ and $F$, $N$ on $AB$ so that $BE \perp AC$, $CF\perp AB$, $\angle ABM = \angle CBM$ and $\angle ACN = \angle BCN$. Prove that $I$ lies on $EF$ if and only if $O$ lies on $MN$.

I honestly can't make any progress. I tried playing around on geogebra but it seems like when I get a configuration where $O$ and $I$ are on the desired lines, $O$ and $I$ stay on these lines as I move $A$ around the circumcenter...

g.kov
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cooldude1212
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    I have an ugly coordinate-based argument showing that the underlying condition is $\cos A = \cos B + \cos C$. This may help lead to a clean proof. – Blue Mar 28 '16 at 07:05
  • oh thats cool, I tried coordinate bashing but gave up due the messy expressions :( how did you do your bash? – cooldude1212 Mar 29 '16 at 05:33
  • I used the unit circle as the circumcircle and set $$A=(1,0) \qquad B=(\cos 2\gamma, \sin 2\gamma) \qquad C = (\cos 2\beta,-\sin 2\beta)$$ where $\beta = \angle B$ and $\gamma = \angle C$. I just let Mathematica do most of the symbol crunching, although I needed to finesse some things along the way. (Mathematica doesn't always simplify the way I like.) Not pretty. In the end, the conditions for $I$ being on $EF$ and $O$ being on $MN$ required some manipulation before they reduced to the common form I gave above. Considering what that form is, I suspect there must be a nice proof somewhere. – Blue Mar 29 '16 at 05:49

3 Answers3

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As the saying goes ... "To someone with a hammer, everything looks like a nail." Here's a solution using a result I proved in another answer:

Extended Ceva's Theorem. Consider points $D_1$, $D_2$, $E_1$, $E_2$, $F_1$, $F_2$ on the (extended) edges of $\triangle ABC$, with $D_i$, $E_i$, $F_i$ on the (extended) edge opposite vertex $A$, $B$, $C$, respectively.

Extended Ceva' Theorem

Lines $\overleftrightarrow{D_1E_2}$, $\overleftrightarrow{E_1F_2}$, $\overleftrightarrow{F_1D_2}$ concur if and only if $$\begin{align} 1 &= \frac{|BD_1|}{|D_1C|} \frac{|CE_1|}{|E_1A|} \frac{|AF_1|}{|F_1B|} + \frac{|D_2C|}{|BD_2|} \frac{|E_2A|}{|CE_2|} \frac{|F_2B|}{|AF_2|} \\[6pt] &+ \frac{|BD_1|}{|D_1C|} \frac{|D_2C|}{|BD_2|} + \frac{|CE_1|}{|E_1A|} \frac{|E_2A|}{|CE_2|} + \frac{|AF_1|}{|F_1B|} \frac{|F_2B|}{|AF_2|} \tag{$\star$} \end{align}$$ Note: The above uses signed lengths, with $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CA}$ indicating the direction of a positively-signed segment on each of the triangle's sides.

For the problem at hand, we will show via $(\star)$ that $I$ lying on $\overleftrightarrow{EF}$, and $O$ lying on $\overleftrightarrow{MN}$ are each equivalent to the common condition $\cos A = \cos B + \cos C$.

enter image description here

In the following, we use the standard definitions $a:=|BC|$, $b:=|CA|$, $c:=|AB|$.

  • The incenter condition amounts to the statement that angle bisectors $\overline{BE}$ and $\overline{CF}$ concur with $\overleftrightarrow{EF}$. This is encoded by $(\star)$ by taking $D_1 = B$ and $E_2 = M$; $E_1 = E$ and $F_2 = F$; $F_1=N$ and $D_2=C$: $$\begin{align} 1 &= \frac{|BB|}{|BC|} \frac{|CE|}{|EA|} \frac{|AN|}{|NB|} + \frac{|CC|}{|BC|} \frac{|MA|}{|CM|} \frac{|FB|}{|AF|} + \frac{|BB|}{|BC|} \frac{|CC|}{|BC|} + \frac{|CE|}{|EA|} \frac{|MA|}{|CM|} + \frac{|AN|}{|NB|} \frac{|FB|}{|AF|} \\[6pt] &= 0+ 0 + 0 + \frac{a\cos C}{c\cos A}\cdot\frac{c}{a} + \frac{b}{c}\cdot\frac{a\cos B}{b\cos A} \end{align}$$ so that $$\cos A = \cos B + \cos C$$

  • The circumcenter condition amounts to the statement that perpendicular bisectors $\overleftrightarrow{B'E'}$ and $\overleftrightarrow{C'F'}$ concur with $\overleftrightarrow{MN}$. This is encoded by $(\star)$ by taking $D_1=B'$ and $E_2=E'$; $E_1=M$ and $F_2=N$; $F_1=F'$ and $D_2=C'$. The result is a little trickier to establish here, so let's define $e := |EE'|$ and $f:=|F'F|$ and make some preliminary calculations (recalling that the distances are signed):$$ \frac{|BB'|}{|B'C|}=-\frac{|EE'|\sec C}{|CE'|\sec C} = -\frac{2e}{b} \qquad \frac{|C'C|}{|BC'|}= -\frac{2f}{c} \qquad \frac{|CE'|}{|E'A|}=\frac{|AF'|}{|F'B|}=1$$ Then $(\star)$ becomes $$\begin{align} 1 &= \frac{|BB'|}{|B'C|} \frac{|CM|}{|MA|} \frac{|AF'|}{|F'B|} + \frac{|C'C|}{|BC'|} \frac{|E'A|}{|CE'|} \frac{|NB|}{|AN|} + \frac{|BB'|}{|B'C|} \frac{|C'C|}{|BC'|} + \frac{|CM|}{|MA|} \frac{|E'A|}{|CE'|} + \frac{|AF'|}{|F'B|} \frac{|NB|}{|AN|} \\ &= -\frac{2e}{b}\cdot\frac{a}{c}\cdot 1 - \frac{2f}{c} \cdot 1 \cdot \frac{a}{b} + \frac{2e}{b} \cdot \frac{2f}{c} + \frac{a}{b} \cdot 1 + 1\cdot \frac{a}{c}\end{align}$$ Expanding each of $e$ and $f$ in two different ways, we have $$\begin{align}bc &= a(b+c-2e-2f) + 4ef \\ &=a\left(b+c-2\left(\tfrac12b-a\cos C\right)-2\left(\tfrac12c-a\cos B\right)\right)+4\left(c\cos A-\tfrac12b\right)\left(b\cos A-\tfrac12c\right) \\ &=2a^2\left(\cos B+\cos C\right)+4bc\cos^2A-2b^2\cos A-2c^2\cos A+b c \\ 0&=a^2\left(\cos B+\cos C\right)-\left(b^2+c^2-2bc\cos A\right)\cos A \\ 0&=a^2\left(\cos B+\cos C\right)-a^2\cos A \end{align}$$ so that, again $$\cos A = \cos B + \cos C$$

This completes the proof. $\square$

(Given how easily the circumcenter case was handled in @James' answer, and how immediately the duality of cases is established in @Karl's answer, I suspect the derivation above could be streamlined a bit. Nevertheless, this is a nice application of the Extended Ceva's Theorem.)

Community
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Blue
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Here is a "synthetic" solution with metric arguments:

We will show both conditions are equivalent with $\cos B +\cos C=\cos A$.

Condition 1: $O$ lies on $MN$. Let $\delta (P, XY)$ be the distance from point $P$ to line $XY$. Notice that since $M,N$ are the feet of the angle bisectors that $\delta (M, AB) + \delta (M, AC) = \delta (M, AB) = \delta (M, BC)$.

Meanwhile, $\delta (N, AB)+\delta (N,AC) = \delta (N,AC)=\delta (N,BC)$.

Since distances from points to lines form linear functions, the function $\delta (P, AB)+\delta (P, AC)-\delta (P, BC)$ must be zero for any point $P\in MN$, because it is zero for $P=M,N$.

Then if $O$ lies on the line, $\delta (P, BC)= R\cos A, \delta (P, AC)=R \cos B, \delta (P, AB)=R\cos C$ which implies $\cos B+\cos C=\cos A$.

Condition 2: $I$ lies on $EF$.

Remark that quadrilateral $BCEF$ is cyclic. Let $X$ be a point on segment $EF$ with $BF=XF$. Note that $\angle BXF = 0.5( \angle BXF + \angle FBX) = 0.5(\angle AFX) = 0.5 \angle ACB = \angle ICB$. This implies that quadrilateral $BXIC$ is cyclic.

As a result, since $\angle IBC=0.5\angle B$, we deduce $\angle IXC = 0.5\angle B$. Then since $\angle XEC = 180-\angle AEF = 180-\angle B$, we deduce that $\triangle XEC$ is isosceles so $XE=EC$.

Then $EF = EX+FX = EC+BF$. But $EF=a \cos A, EC = a \cos C, BF = a \cos B$ so we deduce $\cos A =\cos B + \cos C$ as desired.

Thus the two conditions are equivalent.

James Li
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Easy solution with trilinear coordinates:

I(1;1;1), M(0;1;1), N(1;0;1)

E(0;1,s), F(1,0,r), O(r,s,1) for some r,s because line BE and BO are symmetric with respect to angel bisector from B; (similar CM / CO / C)

Then 

         I on EF

         iff det(I;E;F) = ((1;1;1);(0;1,s);(1,0,r)) = 0

         iff det((r;s;1);(0;1;1);(1;0;1) = 0

         iff det(O;M;N) = 0

         iff O on MN.

This was a problem posed in the german nationwide Highschool contest Bundeswettbewerb Mathematik 2002, second round, Problem 4. Solutions to be found (in german)

https://www.mathe-wettbewerbe.de/bwm/aufgaben

further discussion Bundeswettbewerb Mathematik Die schönsten Aufgaben, p.110

Springer-Verlag Berlin Heidelberg (2016) (ISBN 978-3-662-49539-1)

Karl Fegert
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