Pitching a Tent

Let $a,b,c$ denote the lengths of $BC,CA,AB$ respectively, $h$ denote the height of $\triangle ABC$ from vertex $C$ , and $x$ denote the side length of the square $JKLM$ . Let $(\gamma)$ denote the area of $\gamma$ .

We have the followings $(JKLM)=x^2, (CML)=\dfrac{1}{2} x(h-x),(AMJ)+(BKL)=\dfrac{1}{2}x(c-x).$ Adding them up, we get $(ABC)=x^2+\dfrac{1}{2}x(h-x)+\dfrac{1}{2}x(c-x)=\dfrac 1 2 x(c+h).$ Since $(ABC)=\dfrac 1 2 ch$ we have $x=\dfrac{ch}{c+h}$ .

We can find $(ABC)$ via Heron's formula. Given $a=25,b=26,c=17$ so the half perimeter $s=34$ . Hence $(ABC)=\sqrt{s(s-a)(s-b)(s-c)}=204$ . From $\dfrac 1 2 ch=204$ we have $h=24$ . Thus $x=\dfrac{ch}{c+h}=\dfrac{408}{41}\equiv \dfrac m n\implies m+n=\fbox{449}.$

let a(0,0) b(17,0) giving c(7,24) . So cotA =7/24 and cot B =10/24 If side of the square=x,then AJ =x cotA and KB=x cot B So we have 7/24x +10/24x +x =17 giving x=408/41 and m+n =449

Consider the diagram. By cosine law, we have

$25^2=26^2+17^2-2(26)(17)(\cos A)$ $\implies$ $\angle A\approx 67.38013505^\circ$

$\sin 67.38013505^\circ=\dfrac{h}{26}$ $\implies$ $h=24$

Let $x$ be the side length of the square. Then by substituting to my derived formula, we have

$x=\dfrac{bh}{b+h}=\dfrac{17(24)}{17+24}=\dfrac{408}{41}$

The desired answer is $408+41=$ $\boxed{449}$

First find altitude length from C to AB=CD. In this example, CD divides ABC into two rt. angles triangles ACD a Pythagorean triple 26-24-10,
and CDB, 25-24-7. So CD=24.
Let CM/CA=p. From similar triangles we have,
AB * p=ML=JM=DC * (1 - p). Implies 17p=24 - 24p.
So side of square ML= AB * p=17 * 24/41=m/n.
m+n=408+41=449.

Let x be length of the square.

The area of △ABC = 204.

The height of △ABC to the base AB is 24.

x/(17 - x) = 24 / 17

x = 408/41

and so on >>> good solution