One of the crown jewels of modern geometry

We can place $\triangle ABC$ on a coordinate system, with $A\left( -5,0 \right)$ , $B\left( 9,0 \right)$ and $C\left( 0,12 \right)$ . Then, we have the equations of the lines on which the sides of the triangle are:

$AB:\ y=0, \ \ \ \ \ \text{ }AC: \ 12x-5y+60=0, \ \ \ \ \ \text{ }BC: \ 4x+3y-36=0$ If point $P$ has coordinates $\left( x,y \right)$ , using the distance between a point and a line formula ,

$P{{D}^{2}}+P{{E}^{2}}+P{{F}^{2}}=\dfrac{{{\left( 12x-5y+60 \right)}^{2}}}{{{13}^{2}}}+\dfrac{{{\left( 4x+3y-36 \right)}^{2}}}{{{5}^{2}}}+{{y}^{2}}\coloneqq f\left( x,y \right)$ Taking partial derivatives,

$\begin{aligned} & {{f}_{x}}\left( x,y \right)=\frac{32\left( 394x+33\left( y-12 \right) \right)}{4225} \\ & {{f}_{y}}\left( x,y \right)=\frac{2\left( -25752+528x+6371y \right)}{4225} \\ \end{aligned}$ we find the critical numbers:

$\left. \begin{matrix} {{f}_{x}}\left( x,y \right)=0 \\ {{f}_{y}}\left( x,y \right)=0 \\ \end{matrix} \right\}\Leftrightarrow \left( x,y \right)=\left( \dfrac{198}{295},\dfrac{1176}{295} \right)$ Performing the second partials test we see that we have a minimum, hence these are the coordinates of point $P$ . Consequently, we get $F\left( \dfrac{198}{295},0 \right)$ and $AF=5+\frac{198}{295}=\frac{1673}{295}$ For the answer, $a=1673$ , $b=295$ , thus $a+b=\boxed{1968}$ .

Let $PE = h_a$ , $PD=h_b$ , and $PF=h_c$ . Then the area of $\triangle ABC$ is given by $A = \dfrac {ah_a+bh_b+ch_c}2$ , where $a$ , $b$ , and $c$ are the side lengths opposite angles $A$ , $B$ , and $C$ respectively (not to be confused with answer expression $\dfrac ab$ ). In this case, $13h_b+14h_c+15h_a = 168$ . By Cauchy-Schwarz inequality , we have:

$\begin{aligned} (13h_b+14h_c+15h_a)^2 & \le (13^2 + 14^2 + 15^2) (h_a^2+h_b^2+h_c^2) \\ \implies h_a^2+h_b^2+h_c^2 & \ge \frac {168^2}{13^2 + 14^2 + 15^2} \end{aligned}$

And equality occurs when $\dfrac {h_b}{13} = \dfrac {h_c}{14} = \dfrac {h_a}{15}$ or $h_a = \dfrac {252}{59}$ , $h_b = \dfrac {1092}{295}$ , and $h_c = \dfrac {1176}{295}$ . We also note that:

$AF = \frac 5{12}\left(h_c + \frac 5{13}h_b\right) + \frac {12}{13} h_b = \frac {1673}{295}$

Therefore the required answer is $1673+295 = \boxed{1968}$ .

Place the points so that they are at $A(0, 0)$ , $B(14, 0)$ , $C(5, 12)$ , and $P(p, q)$ .

Then $AC$ is on the line $12x - 5y = 0$ and $CB$ is on the line $4x + 3y - 56 = 0$ , and by the distance from a point to a line equation , $PD^2 = \bigg(\cfrac{12p - 5q}{13}\bigg)^2$ , $PE^2 = \bigg(\cfrac{4p + 3q - 56}{5}\bigg)^2$ , and $PF^2 = q^2$ . That means:

$PE^2 + PF^2 + PD^2$

$= \bigg(\cfrac{4p + 3q - 56}{5}\bigg)^2 + q^2 + \bigg(\cfrac{12p - 5q}{13}\bigg)^2$

$= \cfrac{6304}{4225}p^2 + \cfrac{1056}{4225}pq - \cfrac{448}{25}p + \cfrac{6371}{4225}q^2 - \cfrac{336}{25}q + \cfrac{3136}{25}$

$= \cfrac{8}{832325}(394p + 33q - 2366)^2 + \cfrac{1}{58115}(295q - 1176)^2 + \cfrac{14112}{295}$

which has a minimum of $\cfrac{14112}{295}$ when $394p + 33q - 2366 = 0$ and $295q - 1176 = 0$ .

These two equations solve to $p = \cfrac{1673}{295}$ and $q = \cfrac{1176}{295}$ .

Therefore, $AF = p = \cfrac{1673}{295}$ , so $a = 1673$ , $b = 295$ , and $a + b = \boxed{1968}$ .

By Lagrange's Identity , we have $(a^2+b^2+c^2)(PE^2+PF^2+PD^2)=(a\cdot PE+b\cdot PF+c\cdot PD)^2+(a\cdot PF-b\cdot PE)^2+(a\cdot PD-c\cdot PE)^2+(b\cdot PD-c\cdot PF)^2.$ Since the quantities $a^2+b^2+c^2$ and $a\cdot PE+b\cdot PF+c\cdot PD$ (the latter is twice the area of $\Delta ABC$ ) are fixed, to minimize $PE^2+PF^2+PD^2$ , we must have $\frac{PE}{k}=\frac{PF}{b}=\frac{PD}{c}=k$ where $k$ is some constant.

Therefore, we have $k=\frac{2 A_{\Delta ABC}}{a^2+b^2+c^2}=\frac{84}{295}.$ Now, set up a cartesian coordinate system where $A(0,0), B(14,0), C(5,12)$ are the vertices of the triangle. Let $(x,y)$ be the coordinate of $P$ , we need to locate the $x$ -coordinate which is the length $AF$ . The line $AC$ is given by the equation $AC: 5y-12x=0$ so the distance from point $P$ to line $AC$ is $\frac{|5y-12x|}{13}=k\cdot 13=\frac{1092}{295}$ but we know that $y=k\cdot 14=\frac{1176}{195}$ , substituting it back yields $\boxed{x=\frac{1673}{295}} \ \ (x>0).$

Let $P = (x, y)$

The sum $PF^2 + PE^2 + PD^2 = (PA \cdot u_1)^2 + (PB \cdot u_2)^2 + (PC \cdot u_3)^2 = (P - A)^T u_1 u_1^T (P - A) + (P - B)^T u_2 u_2^T + (P - C)^T u_3 u_3^T (P - C)$

where $u_1 , u_2, u_3$ are unit vectors orthogonal to $AB, BC, CA$ respectively.

Thus $u_1 = (0, 1) , u_2 = (\frac{4}{5}, \frac{3}{5}) , u_3 = ( \frac{12}{13}, -\frac{5}{13} )$

The expression for the sum of the squared distances can be further simplifed as follows

$PF^2 + PE^2 + PD^2 = P^T Q P + b^T P + c = f(P)$

where $Q = u_1 u_1^T + u_2 u_2^T + u_3 u_3^T , b = - 2 ( u_1 u_1^T A + u_2 u_2^T B + u_3 u_3^T C ), c = A^T u_1 u_1^T A + B^T u_2 u_2^T B + C^T u_3 u_3^T C$

Since $u_1 u_1^T, u_2 u_2^T , u_3 u_3^T$ are positive semi-definite, and the vectors $u_1 , u_2 , u_3$ are not collinear, then $Q$ is positive definite, then our objective function $f(P)$ attains its minimum at $P^* = -\dfrac{1}{2} Q^{-1} b$ . This is easy to show by completing the square.

The rest is numerical computations, which are as follows:

First, we have $A = (0, 0) , B = (14, 0), C = (5, 12)$

Secondly,

$u_1 u_1^T = \begin{bmatrix} 0 && 0 \\ 0 && 1 \end{bmatrix}$

$u_2 u_2^T =\dfrac{1}{25} \begin{bmatrix} 16 && 12 \\ 12 && 9 \end{bmatrix}$

$u_3 u_3^T = \dfrac{1}{169} \begin{bmatrix} 144 && -60 \\ -60 && 25 \end{bmatrix}$

Therefore,

$Q = \dfrac{1}{4225} \displaystyle \begin{bmatrix} 6304 && 528 \\ 528 && 6371 \end{bmatrix}$

$b =(-2) (\dfrac{224}{25} , \dfrac{168}{25} )$

Hence,

$Q^{-1} = \dfrac{1}{9440} \displaystyle \begin{bmatrix} 6371 && -528 \\ -528 && 6304 \end{bmatrix}$

And,

$P^{*} = -\dfrac{1}{2} Q^{-1} b = (\dfrac{1673}{295} , \dfrac{1176}{295} )$

Hence the required distance $AF = \dfrac{1673}{295}$ , and therefore the answer is $1673 + 295 = \boxed{1968}$ .

Looking at the picture, it is clear that fixing the distances AD and BE will uniquely determine the location of point P, which is at the intersection.

Define $AD=r, BE=s, DP=t, EP=u, FP=y, AF=x$ The outline of this solution is

• establish the angles
• to express t, u and y in terms of r and s
• find r and s by setting $d(u^2+t^2+y^2)/dr$ and $d(u^2+t^2+y^2)/ds$ to 0
• find AF

Establish the angles

Let the interior angles at $A, B, C$ be $α,β,γ$ respectively

Apply the cosine rule: $\cos α = \frac{13^2+14^2-15^2}{2\cdot 13 \cdot 14}=\frac{140}{364}=\frac{5}{13}$ ; $\sin α = \sqrt{1-\cos^2α}= \frac{12}{13}$ This way we get

Express t, u and y in terms of r and s

Setting the origin in A, use the sine and cosine values to express coordinates in terms of r,s,t and u: $D=(\frac{5}{13}r,\frac{12}{13}r)$ $E=(14-\frac{3}{5}s,\frac{4}{5}s)$ $P=(x,y)=(\frac{5r+12t}{13}, \frac{12r-5t}{13})=(14-\frac{3s+4u}{5},\frac{4s-3u}{5})$

Using the two expressions for x and the two expressions for y, we can eliminate u and t. $65x=25r+60t=910-39s-52u$ $65y=60r-25t=52s-39u$ Combine these to express t and u in terms of r and s: $t= \frac{546+33r-65s}{56}, u=\frac{350-65r+33s}{56} \text{ and } y=\frac{4s-3u}{5}=\frac{-210+39r+25s}{56}$

Find r and s

P is such that it minimizes $Q(r,s)= t^2+u^2+y^2 = ( (546+33r-65s)^2+(350-65r+33s)^2+(-210+39r+25s)^2)/56^2$ $56^2Q = 464716+6835r^2+5939s^2+-25844r -58380s -6630rs$

$dQ/dr=0 \Rightarrow 6835r-12922 -3315s = 0$ $dQ/ds=0 \Rightarrow 5939s -29190 -3315r=0$ these solve to $r=1729/295, s= 483/59$

Find AF

$t= \frac{546+33× 1729/295 -65×483/59}{56}= \frac{1092}{295}$ $u=\frac{350-65×1729/295+33× 483/59}{56} = \frac{252}{59}$ $y=\frac{-210+39×1729/295+25× 483/59}{56}= \frac{1176}{295}$ $x=\frac{5×1729/295 +12× 1092/295}{13}= \frac{1673}{295}$ Since $AF=x$ we should submit $1673+295=\boxed{1968}$

The minimum value for $t^2+u^2+y^2= \frac{14112}{295}$