Dgdrxrt

In $triangle ABC$, I is the incenter. Area of $triangle IBC = 28$, area of $triangle ICA = 30$ and area of $triangle IAB = 26$. Find $AC^2 − AB^2$.

Here is a sketch that I drew:


From the given areas ,
$r*AC=60$
$r*AB=52$
$r*BC=56$

So that I've three equations with 4 unknowns .
Also,
$A=rs$
$2A=168=r(AB+BC+AC)$

But this can't be used as it can be derived from the other three equations.
So, what should I do? Any hints are apreciated.
(This is not class-homework , I'm solving sample questions for a competitive exam )

Note that $b=frac15a14$, $c=frac13a14$. Thus, $s=frac14a+13a+15a2cdot14=frac3a2$. So by Heron's formula, $$84=sqrtfrac3a^24cdot frac3a7cdot frac4a7=frac6a^214$$

Thus, $a^2=14^2$, or $a=14$. Now you can go on from here.

we note $AC=b,BC=a,AB=c$, we will find $b^2-c^2$. We have
$$ar=56,br=60,cr=52.$$
then we have the proportion of $a$ and $b,c$:
$$b=frac6056a,c=frac5256a$$
Then we have the equality:
$$a=frac56112(b+c)=frac12(b+c)$$
$$a=frac568(b-c)=7(b-c)$$
Now we should recall the formula of triangle's area:
$S=sqrtp(p-a)(p-b)(p-c)$, where $p=fraca+b+c2$
Then we have
$$84=sqrtfraca+b+c2cdotfracb+c-a2cdotfraca+c-b2cdotfraca+b-c2\=frac14sqrt[(b+c)^2-a^2][a^2-(b-c)^2]\=frac14sqrt[(b+c)^2-frac14(b+c)^2][49(b-c)^2-(b-c)^2]\=frac14sqrtfrac34(b+c)^2cdot48(b-c)^2\frac32|b^2-c^2|\=frac32(b^2-c^2)$$(because $b>c$).

then we have what we want.

A remark on the solvability of the question: you are given $ra=56,rb=60,rc=52$ (with the usual notations $r$ the inradius, $a=BC$, etc). Therefore you can deduce that the triangle $ABC$ is similar to the triangle $Delta=(56,60,52)$ with scaling ratio $r$.

You find the area of $ABC$ by summing up the areas of the three triangles and use the fact that $r^2 Area(ABC)=Area(Delta)$ (you can calculate $Area(Delta)$). This gives you $r$, and therefore you can find $a,b,c$.