sábado, 30 de julio de 2016

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 $CF = CE$.


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, $CF=CE$.

jueves, 28 de julio de 2016

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.

miércoles, 27 de julio de 2016