Let ABC be a right-angled isosceles triangle where AB = BC = a. Assume that C is its centroid and I is its incenter. Find, in terms of a, the distance between C and I.
Answer : $CI= \frac{{a \cdot (3\sqrt{2}-4})}{12}$
How to find it ?
[Math] Distance between centroid and incenter in a right-angled isosceles triangle
geometrytrigonometry
Related Solutions
- in $\bigtriangleup$ABC AB=AC
- WE take AD$\perp$BC.clearly BD=DC & $\angle$BAD=$\angle$DAC
- so clearly the circumcentre;orthocentre;incentre and centroid - all of them lie on the line AD
- SO that is prooved
The side lengths are $a=25,b=39,c=56$ by the pythagorean theorem and the area is $\Delta=420$ by the Heron's formula or the shoelace formula, hence $r=7$. Since $h_c=\frac{2\Delta}{c}$, the length of the height relative to $C$ is $15$, hence the distance of $G$ from $AB$ is just $5$.
It is well known that $I=\frac{aA+bB+cC}{a+b+c}$ and $AI^2=bc-4rR=bc-\frac{2abc}{a+b+c}$, hence $IG^2$ can also be computed through the parallel axis theorem. Let we assume that $A$ has mass $a$, $B$ has mass $b$ and $C$ has mass $c$. With such assumptions, $I$ is the center of mass of $S=\{A,B,C\}$ and the moment of inertia of $S$ around $I$ is given by: $$ M_I = a AI^2 + b BI^2 + c CI^2 = 3abc - 4(a+b+c) rR = abc $$ while the moment of inertia of $S$ around $G$ is given by: $$ M_G = a AG^2 + b BG^2 + c CG^2 = \frac{1}{9}\sum_{cyc}a(2b^2+2c^2-a^2) $$ and by the parallel axis theorem $ M_G = (a+b+c) IG^2 + M_I$, hence:
$$ IG^2 = \frac{M_G-M_I}{a+b+c} = \boxed{\frac{\frac{2}{3}(a+b+c)^3-\frac{5}{3}(a^3+b^3+c^3)-13abc}{9(a+b+c)}}.$$
A difficult problem has just born: Prove that if $a,b,c$ are the side lengths of a triangle, $$ 2(a+b+c)^3 \geq 5 (a^3+b^3+c^3) + 39abc. $$
Tricky solution: $IG^2\geq 0$, and equality is achieved only when $I\equiv G$, i.e. when $a=b=c$.
Best Answer
If Coordinate Geometry is allowed, WLOG we can assume $\displaystyle B(0,0),A(a,0), C(0,a)$
So, the equation of $AC$ will be $\displaystyle\frac xa+\frac ya=1\iff x+y-a=0$
So, the centroid will be $\displaystyle C\left(\frac a3,\frac a3\right)$
Now, if $I(p,q)$ then we have the perpendicular distance of $I$ from $AB,BC,CA$ will be same
$\displaystyle\implies |p|=|q|=\frac{|p+q-a|}{\sqrt{1^2+2^2}} $
If $\displaystyle p,q>0,p=q=\frac{2p-a}{\sqrt2}\implies (2-\sqrt2)p=a\iff p=\frac a{2+\sqrt2}$
Now, $IC=|\sqrt{(p-a)^2+(q-a)^2}|=\sqrt2|p-a|$
Can you take it from here?