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Source: IberoAmerican Mathematical Olympiad 2004, Problem 5

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#1 h Sep 23, 2004, 10:38 AM • 2 Y

Y by Adventure10, Mango247

Given a scalene triangle $ABC$ . Let $A'$ , $B'$ , $C'$ be the points where the internal bisectors of the angles $CAB$ , $ABC$ , $BCA$ meet the sides $BC$ , $CA$ , $AB$ , respectively. Let the line $BC$ meet the perpendicular bisector of $AA'$ at $A''$ . Let the line $CA$ meet the perpendicular bisector of $BB'$ at $B'$ . Let the line $AB$ meet the perpendicular bisector of $CC'$ at $C''$ . Prove that $A''$ , $B''$ and $C''$ are collinear.

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#2 h Sep 23, 2004, 11:30 AM • 2 Y

Y by Adventure10, Mango247

This is an old problem from Igor Sharygin (exercise 3 in his Kvant article "Теоремы Чевы и Менелая"). It is the starting point of a theory about some triangle centers which I later named "Sharygin points".

Here is a solution of the problem:

Let the tangents to the circumcircle of triangle ABC at the points A, B, C meet the lines BC, CA, AB at the points X, Y, Z, respectively. It is well-known that these points X, Y, Z are collinear. Now, we will prove that A" = X, B" = Y and C" = Z; this will clearly solve the problem.

Since the point B" lies on the perpendicular bisector of the segment BB', the triangle BB"B' is isosceles. Thus, < B"BB' = < B"B'B. Hence,

< B"BA = < B"BB' - < ABB' = < B"B'B - < ABB' = < AB'B - < ABB'
= (180 - < ABB' - < BAB') - < ABB' = 180 - 2 < ABB' - < BAB'
= 180 - 2 B/2 - A = 180 - B - A = C = < BCA.

On the other hand, since the line BY is tangent to the circumcircle of triangle ABC, we have < YBA = < BCA (by the secant-tangent angle theorem). Together with < B"BA = < BCA, this implies < YBA = < B"BA, and hence the point Y lies on the line BB". Since we also know that both points Y and B" lie on the line CA, it follows that B" = Y. Similarly, C" = Z and A" = X. This completes the solution.

Darij

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#3 h May 23, 2007, 11:18 AM • 3 Y

Y by Adventure10, Mango247, Mango247

Consider that $\angle CBA > \angle BCA$ now we have $\angle A''AA'=\angle A''A'A=\angle BAC/2+\angle ACB$
$\longrightarrow \angle A''AB=\angle ACB$ , so $\triangle A''AB$ and $\triangle A''AC$ are similiar and we have :
$\frac{A''B}{A''A}=\frac{A''A}{A''C}=\frac{AB}{AC}\longrightarrow$ $\frac{A''B}{A''C}=\frac{AB^{2}}{AC^{2}}$ .
similiary we get that : $\frac{B''C}{B''A}=\frac{BC^{2}}{BA^{2}}$ and $\frac{C''A}{C''B}=\frac{CA^{2}}{CB^{2}}$ .
now we can check the Menelaus' theorem easily !

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#4 h May 24, 2007, 7:54 AM • 2 Y

Y by Adventure10, Mango247

This nice, easy and old problem belongs to the clasical synthetical geometry.

Proof. Denote $X\in AA\cap BC$ . Observe that $\{\begin{array}{c}m(\widehat{XAA'})=m(\widehat{XAB})+m(\widehat{BAA'})=C+\frac{A}{2}\\\ m(\widehat{XA'A})=m(\widehat{A'AC})+m(\widehat{A'CA})=\frac{A}{2}+C\end{array}$ $\implies$

$\widehat{XAA'}\equiv\widehat{XA'A}$ $\implies$ $\triangle AXA'$ is isosceles. In conclusion $X\equiv A''$ a.s.o.

This post has been edited 1 time. Last edited by Virgil Nicula, Sep 12, 2013, 5:33 PM

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#5 h May 25, 2007, 6:23 PM • 2 Y

Y by Adventure10, Mango247

Well i tried an ugly vectorial proof. Sorry for eventual mistakes, i am not into the style.

Click to reveal hidden text

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#6 h Jan 1, 2009, 2:23 PM • 2 Y

Y by Adventure10, Mango247

Approach by mihai miculita:

Let $A_1; B_1; C_1$ the intersections of external bisectors of angles $A; B; C$ with the opposite sides. The circles of diametres $[A'A_1], [B'B_1], [C'C_1]$ is the Appollonius circles of $\triangle{ABC}$ and the points $A", B", C"$ is the Appolonian circles centres $\Rightarrow$ the points $A", B", C"$ coliniar!

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#7 h Jan 1, 2009, 10:34 PM • 1 Y

Y by Adventure10

Also, it has been posted as An interesting line of a triangle, by nayel.

Kostas Vittas.

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#8 h Sep 8, 2013, 6:51 PM • 2 Y

Y by Adventure10, Mango247

Sorry for reviving, but...

Let the $A$ -tangent to $(ABC)$ meet $\overleftrightarrow{BC}$ in $A^{*}$ , then $\measuredangle A^{*}A'A=\measuredangle A'AC+\measuredangle A'CA=\measuredangle BAA'+\measuredangle BAA^{*}=\measuredangle A^{*}AA'$ , so $\Delta A^{*}AA'$ is $A^{*}$ -isosceles thus perpendicular bisector of $\overline{AA'}$ passes through $A^{*}$ . This shows that $A^{*}=A''$ .
As we know, $A^{*}$ , $B^{*}$ and $C^{*}$ form the Lemoine's line of $\Delta ABC$ , so $A''$ , $B''$ and $C''$ are colinear.

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#10 h Sep 13, 2016, 10:35 PM • 1 Y

Y by Adventure10

Note that $\angle A''AB = \angle A''AA' - \angle A/2 = 180^{\circ}-\angle B - \angle A/2 - \angle A/2 = \angle C$ so $A''A$ is tangent to $(ABC)$ at $A$ .

Let $X,Y,Z$ be the points on $BC,CA,AB$ respectively such that $AX,BY,CZ$ are symmedians. Let $T$ be the intersection of the tangents to $(ABC)$ at $B,C$ . Then $A'' \in BC$ the polar of $T$ so $T$ is in the polar of $A''$ . Similarly $A$ lies in the polar of $A''$ because $A''A$ is tangent to $(ABC)$ .

Therefore $AT$ is the polar of $A''$ and it is well known that $AT$ is the symmedian so $X$ is also in the polar of $A''$ . Thus by a well known lemma we have $(B,C;X,A'') = -1$ .

We know that $AX,BY,CZ$ are concurrent at the symmedian point of $\triangle ABC$ so $YZ \cap BC = A_1$ satisfies $(B,C;X,A_1) = -1$ . Therefore $A'' = YZ \cap BC$ and similarly $B'' = XZ \cap AC, C'' = XY \cap AB$ .

Since $AX,BY,CZ$ concur we know that triangles $ABC$ and $XYZ$ are perspective meaning $AB \cap XY, BC \cap YZ, CA \cap XZ = A'',B'',C''$ are collinear as desired.

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#11 h Sep 14, 2016, 10:02 PM • 2 Y

Y by Adventure10, Mango247

See P5 from here

This post has been edited 1 time. Last edited by Virgil Nicula, Sep 15, 2016, 8:47 PM

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