Let $ABC$ be triangle with sides that are not equal. Find point $X$ on $BC$ From following conditions.

Question

Let $ABC$ be triangle with sides that are not equal. Find point $X$ on $BC$ such that $$\frac{\text{area}\ \triangle{ABX}}{\text{area}\ \triangle{ACX}}=\frac{\text{perimeter}{\ \triangle{ABX}}}{\text{perimeter}{\ \triangle {ACX}}}$$

I am not getting how to start with, Clearly $ABC$ triangle is scalene but what about other triangle is it necessary they should be scalene too? Please help me to solve this. Thanks in advance

Answer 1

Hint: It makes sense to rewrite this as $$\frac{\text{area}\ \triangle{ABX}}{\text{perimeter}{\ \triangle{ABX}}}=\frac{\text{area}\ \triangle{ACX}}{\text{perimeter}{\ \triangle {ACX}}},$$ since that way each side depends only on one triangle. Now, what do you know about the ratio of area to perimeter? Does it equal any other quantity in a triangle that you know about?

Answer 2

You can solve it using the help of Equal Incircles Theorem. Here I am telling you steps to draw and find point $X$.

You have a scalene triangle $ABC$ that is given. We have to find a point $X$ on BC that will divide the triangle $ABC$ into two with equal inradius (as the ratio of perimeter and area are same).

Step 1: Find the height from point A to BC. Say, $h$.

Step 2: Find its inradius. Say, $r$.

Step 3: Find inradius of the two new triangles $ABX$ and $ACX$ using Equal Incircles theorem (please refer to https://www.cut-the-knot.org/triangle/EqualIncirclesTheorem.shtml),

$(1 - \dfrac{2r_1}{h})^2 = 1 - \dfrac{2r}{h} \,$ where $r_1$ is the inradius of two new triangles.

Step 4: As the incircles of triangles $ABX$ and $ACX$ will be both touching line $BC$, draw a line $DE$ parallel to $BC$ at distance $r_1$. Wherever bisector of $\angle ABC$ and $\angle BCA$ intersect line $DE$ ($F$ and $G$) are the incenters of triangles $ABX$ and $ACX$.

Step 5: Draw a circle with radius $r_1$ at $F$ or at $G$. Then draw a tangent to this circle from point $A$. The point where the tangent intersects $BC$ is the point $X$ you want.

Answer 3

Obviously, the stated condition means that $\triangle ABX$ and $\triangle AXC$ must have the same radius of the inscribed circle. That means $AX$ must be the incircle bisector of $\triangle ABC$, see Incircle bisectors and related measures.

For $A$, $B$, $C$ ordered in positive (counterclockwise) direction, $|BC|=a$, $|AC|=b$, $|AB|=c$, using expressions given in the link above, we can find out that \begin{align}|BX|&=\tfrac12\,a-\frac{b-c}{2a}\,\Big(b+c-\sqrt{(b+c)^2-a^2}\Big).\end{align}