# Conic in Mixtilinear Incircles

### Problem

The six points of intersection of the mixtilinear incircles with the sides of a triangle lie on a conic. ### Solution

Let the points of intersection be $A_b,$ $A_c,$ etc. as shown in the diagram.$ According to Carnot's Theorem for Conics, the six points will lie on a conic, provided $\displaystyle \big(\frac{AC_a}{BC_a}\cdot\frac{AC_b}{BC_b}\big)\cdot\big(\frac{BA_b}{CA_b}\cdot\frac{BA_c}{CA_c}\big)\cdot\big(\frac{CB_c}{AB_c}\cdot\frac{CB_a}{AB_a}\big)=1.$ Let$K$and$M$be the points of tangency of two of the mixtilinear incircles with side$AC.$ Then, by the Power of a Point Theorem,$AC_a\cdot AC_b=AK^{2}$and$CA_b\cdot CA_c=CM^{2}.$Now, we know that$AK=\displaystyle\frac{ab}{p}$and$CM=\displaystyle\frac{bc}{p},$where$a,b,c$are the side lengths opposite vertices$A,B,C$and$p=(a+b+c)/2\$ the semiperimeter. The other elements in Carnot's criterion are defined cyclically, and this is why the condition holds: upon substitution all the terms cancel out.

$\displaystyle \bigg(\frac{ab/p}{ab/p}\cdot\frac{bc/p}{bc/p}\cdot\frac{ac/p}{ac/p}\bigg)^{2}=1.$

### Acknowledgment

The problem has been conjectured by Emmanuel Antonio José García; the solution at the community blog of the artofproblemsolving site is due to Ivan Zelich. ### Mixtilinear Incircles

• Mixtilinear Circles and Concurrence
• Sangakus with a Mixtilinear Circle
• Radius and Construction of a Mixtilinear Circle
• Sangaku with Three Mixtilinear Circles
• Construction and Properties of Mixtilinear Incircles
• Construction and Properties of Mixtilinear Incircles 2
• A Sangaku: Two Unrelated Circles
• 