3 consecutive integer.

Geometry Level 3

In $\triangle ABC,$

the incircle touches sides $BC, CA, AB$ at points $D, E, F,$ respectively;
the inradius has length 4;
the lengths of $BD, CE, AF$ are consecutive integers;
the side lengths of the triangle are $X, Y, Z.$

Find $X+Y-Z.$

Note: $Z$ is the length of side $AC.$

The answer is 12.

1 solution

Eduardo Ag_182
Mar 30, 2018

$BD = a$
$DC = a+1$
$CE = a+1$
$EA = a+2$
$AF = a+2$
$FB = a$
$X= AB = 2a+2$
$Y= BC = 2a+1$
$Z= AC = 2a+3$

Semiperimeter. $S= 3a+3$

Area. $A=R*S= 4(3a+3) = 12a+12$

Area, Heron's formula

$A=\sqrt{S(S-X)(S-Y)(S-Z)}$
$A=\sqrt{(3a+3)(3a+3-2a-2)(3a+3-2a-1)(3a+3-2a-3)}$
$A=\sqrt{(3a+3)(a+1)(a+2)(a)}$
$A=\sqrt{3a^4+12a^3+15a^2+6a}$

then:

$12a+12= A =\sqrt{3a^4+12a^3+15a^2+6a}$
$(12a+12)^2=3a^4+12a^3+15a^2+6a$

simplifying:

$a^4+4a^3-43a^2-94a-48=0$

Roots are:

$a=6$
$a=-8$
$a=-1 ,duplicate$

So $a=6$

$X=14, Y= 13, Z=15$

$X+Y-Z = 2a = 2(6) = 12$