The Outer/Inner Garcia Triangle

In euclidean geometry, the Fuhrmann triangle and the Hexyl triangle (and others) are two special triangles with a plethora of surprising properties. On April, 2014, I introduced two triangles, namely, the Outer Garcia triangle and the Inner Garcia triangle which, as it regards to intersting properties, seem to compete very well with the two triangles cited above. 

Definitions:

Outer-Garcia Triangle: it is the triangle obtained by reflecting the incenter around the midpoints of the reference triangle. This is triangle $A'B'C'$ in the animation.

Inner-Garcia Triangle: it is the triangle obtained by reflecting the vertices of the Outer-Garcia triangle around the sides of the reference triangle. Also, It can be obtained by just reflecting the incenter around the perpendicular bisectors of the reference triangle. This is triangle $A''B''C''$ in the animation.

The constructions and many of the properties associated to these triangles can be found in the Encyclopedia of Triangle Centers, part 4, with notation $X(5587)$.

There are also cubics associated to them which you can find in the website of Bernard Gibert. See, for example, the Spieker-Schiffler Cubic.

Related topic:
A Note on Reflections
Reflections on Poncelet’s Pencil

Concyclicity Associated to a Radical Axis

Consider two circles with centers $A$ and $B$. Call $A'$, $A''$ the two intersections of the line $AB$ and the circle centered at $A$. Similarly, construct $B'$, $B''$. Call $X$ and $Y$ the intersections of circles $(A)$, $(B)$ with another circle passing through $A''$, $B''$. Let another circle passing through $A''$, $B''$ be intersected by both circles $(A)$, $(B)$ at $Z$, $W$, respectively.

Prove that $XYWZ$ is cyclic.

Congruent Segments Associated to Tangent Lines.

Let $A$ and $B$ be the centers two circles $(A)$, $(B)$. From $A$, draw a tangent to $(B)$ at $C$. From $B$, draw two tangents to $(A)$ in $D$, $E$. From $C$, draw $CD$ such that intersect $(A)$ again in $F$. Similarly, from $C$, draw $CE$ such that intersect $(A)$ in $I$.

Prove $CF = CE$.

Proof.

Since $BD$ and $AC$ are tangent lines it follows that $\angle{BDA}=\angle{ACB}=\angle{AEB}=90^\circ$, hence $A$, $B$, $C$, $D$, $E$ are concyclic. Now, see that $\angle{DCE}=\angle{DBE}$ and $\angle{ACE}=\angle{ABE}$. $\angle{ABE}=\frac{\angle{DBE}}{2}=\frac{\angle{DCE}}{2}$, then, segment $AC$ is an angle bisector of $\angle{DCE}$. Focus on $\triangle{ACF}$. $\angle{CDB}=\angle{CEB}$. $\angle{FDA}=\angle{DFA}=180^\circ-(90^\circ+\angle{CDB})=90^\circ-\angle{CDB}$.$\angle{FAC}= 180^\circ-\angle{DCA}-(90^\circ-\angle{CDB})=90^\circ+\angle{CDB}-\angle{DCA}$. Focus on $\triangle{ACE}$. $\angle{EAC}=180^\circ-\angle{DCA}-(90^\circ-\angle{CDB})$. $\angle{EAC}=90^\circ+\angle{CDB}-\angle{DCA}$. Hence $\angle{FAC}=\angle{EAC}$, and $\triangle{ACF}$, $\triangle{ACE}$ are congruent. Thus, $CF=CE$.

On a constant associated to equilateral triangle and its generalization

I guess you are familiar with the result described as follows:

If $ABC$ is an equilateral triangle, and $P$ is any point on the incircle of $\triangle{ABC}$, then $AP^2 + B^2 + CP^2$ is constant. Click here to read more.

I was wondering whether this can be true for any regular polygon and found it to be true too. In the website it is mentioned that the result holds for any circle with center at the centroid of the triangle, but it does not mention whether the result holds for any other polygon. Although it seems a natural question, I have not seen any reference so far.

Click here for the generalization.

Relationship Between Inradius and Exradii

Prove $\frac{r}{r_a}+\frac{r}{r_b}+\frac{r}{r_c} = 1$

I am grateful to Angel Rafael Mejia for the reference:

Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.

The Three Square Geometry Problem

This is my solution of the famous Three Square Geometry Problem. 

Click on the image to have a better view.

A Perspector Associated to Cevians

This is an old open problem by  Oai Thanh Đào. Here I give a proof.

Consider a triangle ABC. Let D be a point inside triangle ABC. Let EFG be the cevian triangle of point D. On segments BF, CF, CG, AG, AE, BE erect similar isosceles triangles BFH, CFI, CGJ, AGK, AEL and BEM, respectively. Let N be the intersection of HI, LM. Similarly, let O be the intersection of HI, KJ and P the intersection of KJ, ML.  Then, AP, BN, CO concur. 

Proof (click on the image to have a better view).

Related topics:

A family of perspectors associated to cevians