A Unique Set of Cevians (almost)

Let $AD=DB=CE=x$ . By generalized angle bisector theorem , we have:

$\begin{aligned} \frac {BD}{AD} & = \frac {BC \cdot \sin \angle BCD}{CA \cdot \sin \angle ACD} \\ \frac {14-x}x & = \frac {15 \sin \angle BCD}{13 \sin \angle ACD} \\ \implies \frac {\sin \angle BCD}{\sin \angle ACD} & = \frac {13(14-x)}{15x} \end{aligned}$

Similarly,

$\begin{aligned} \frac {YF}{AF} & = \frac {CF \cdot \blue{\sin \angle BCD}}{CA \cdot \blue{\sin \angle ACD}} \\ \frac {AF-AY}{AY} & = \frac {(15-x)\cdot \blue{13(14-x)}}{13\cdot \blue{15x}} \\ \frac {AF}{AY} - 1 & = \frac {(14-x)(15-x)}{15x} \\ \implies AY & = \frac {15x \cdot AF}{(14-x)(15-x)+15x} \end{aligned}$

We need to find $AF$ and $x$ to find $AY$ . To find $AF$ , we need to first find $x$ and we can use Ceva's theorem as follows.

$\begin{aligned} \frac {CE}{EA} \cdot \frac {AD}{DB} \cdot \frac {BF}{FC} & = 1 \\ \frac x{13-x} \cdot \frac x{14-x} \cdot \frac x{15-x} & = 1 \\ x^3 - (13-x)(14-x)(15-x) & = 0 \\ \implies x & \approx 6.976109308 \end{aligned}$

To find $AF$ , we use cosine rule . First $14^2=13^2+15^2 - 2(13)(15)\cos C \implies \cos C = \dfrac {13^2+15^2-14^2}{2(13)(15)} = \dfrac {33}{65}$ and

$\begin{aligned} AF^2 & = 13^2 + (15-x)^2 - 2(13)(15-x)\cos C \approx 127.4674647 \\ \implies AF & \approx 11.29014901 \end{aligned}$

Therefore, $AY = \dfrac {15x \cdot AF}{(14-x)(15-x)+15x} \approx 7.337984571 \implies \lfloor 10000\cdot AY\rfloor = \boxed{73379}$ .

The cevians are concurrent at Y, the (first) Yff point for $\triangle ABC$ . They are constructed by marking off a distance $x$ , counter-clockwise from each vertex, where $x$ is the solution to the cubic equation $(a - x)(b - x)(c - x) = x^3$ . A second Yff point (named for Peter Yff) can be found by measuring clockwise. $x$ is then used to compute trilinear coordinates , $\alpha:\beta:\gamma$ for $Y$ .

$\displaystyle \alpha = \frac{1}{a} \sqrt[3]{\frac{c-x}{b-x}} ,\hspace{1cm} \beta = \frac{1}{b} \sqrt[3]{\frac{a-x}{c-x}},\hspace{1cm} \gamma = \frac{1}{c} \sqrt[3]{\frac{b-x}{a-x}}$

Solving the cubic , $x \approx 6.977611$

And the trilinears are $\approx 0.070169 :0.080413: 0.064919$

I converted to cartesian coordinates to compute $AY$ , assigning $A = (0,0), B = (14,0), C = (5,12)$ which results in $Y \approx (6.37879, 3.62727)$

So by Pythagoras, $AY = \sqrt{ 6.37879^2 + 3.62727^2} \approx 73379.8$ and $10000{AY} = \fbox{73379}$

More information on the Yff triangle here.

Three steps – three theorems: Ceva’s , Stewart’s , Menelaus’s .

Step 1 : to find $x=AD=BF=CE$
By Ceva’s theorem on $\triangle ABC$ we have $\begin{aligned} \frac{AD}{DB}\cdot \frac{BF}{FC}\cdot \frac{CE}{EA}=1 & \Rightarrow \frac{x}{14-x}\cdot \frac{x}{15-x}\cdot \frac{x}{13-x}=1 \\ & \Rightarrow {{x}^{3}}=\left( 13-x \right)\left( 14-x \right)\left( 15-x \right) \\ & \Rightarrow 2{{x}^{3}}-42{{x}^{2}}+587x-2730=0 \\ \end{aligned}$ which solves to $x=7+\frac{\sqrt[3]{-63+\sqrt{150926511}}}{\sqrt[3]{36}}-\frac{293}{\sqrt[3]{6\left( -63+\sqrt{150926511} \right)}}\Rightarrow x\approx \text{6}\text{.9761093080953896}$

Step 2 : to express $AF$ in terms of $x$
By Stewart’s theorem on $\triangle ABC$ we have $\begin{aligned} & A{{F}^{2}}=\frac{A{{C}^{2}}\cdot BF+A{{B}^{2}}\cdot FC}{BF+FC}-BF\cdot FC \\ & =\frac{{{13}^{2}}\cdot x+{{14}^{2}}\cdot \left( 15-x \right)}{15}-x\cdot \left( 15-x \right) \\ & ={{x}^{2}}-\frac{84}{5}x+196 \\ \end{aligned}$ Thus, $AF=\sqrt{{{x}^{2}}-\frac{84}{5}x+196}$

Step 3 : to find $AY$
By Menelaus’s theorem on $\triangle ABC$ intersected by a line at points $Y$ , $C$ and $D$ we have $\begin{aligned} \frac{AY}{YF}\cdot \frac{FC}{CB}\cdot \frac{BD}{DA}=1 & \Rightarrow \frac{AY}{YF}=\frac{CB}{FC}\cdot \frac{DA}{DA} \\ & \Rightarrow \frac{AY}{YF}=\frac{15x}{\left( 15-x \right)\left( 14-x \right)} \\ & \Rightarrow \frac{AY}{YF}=\frac{15\left( 13-x \right)}{{{x}^{2}}}\ \ \ \ \ \text{ from the equation in step 1} \\ & \Rightarrow AY=\frac{15\left( 13-x \right)}{{{x}^{2}}}\left( AF-AY \right) \\ & \Rightarrow AY\cdot \left( 1+\frac{15\left( 13-x \right)}{{{x}^{2}}} \right)=\frac{15\left( 13-x \right)}{{{x}^{2}}}\cdot \sqrt{{{x}^{2}}-\frac{84}{5}x+196} \\ & \Rightarrow AY=\frac{3\sqrt{5}\left( 13-x \right)\sqrt{5{{x}^{2}}-84x+980}}{{{x}^{2}}-15x+195} \\ \end{aligned}$

Substituting the value of $x$ we found in step 1 we get $AY\approx 7.33798457062463468$ .

For the answer, $\left\lfloor 10000\cdot AY \right\rfloor =\boxed{73379}$ .