Dynamic Geometry: P92

Let the large and small triangles be $ABC$ and $CDE$ respectively, where $B$ , $C$ , and $D$ are fixed tangent points of the two circles with the base line and with each other; $ST$ through $C$ be parallel to $BD$ , $CN$ be perpendicular to $BD$ : and the centers of the small and large circles be $O$ and $P$ respectively.

We note that $OC=OD=1$ and the tangent of $OP$ is $\frac 34$ . Then $DN = \frac 45$ , $CN=1+\frac 35 = \frac 85$ , by Pythagorean theorem , $\blue{CD} = \sqrt{CN^2+DN^2} = \blue{\frac 4{\sqrt 5}}$ , $BN=BD-DN = 4-\frac 45 = \frac {16}5$ , and $\blue{BC} = \sqrt{BN^2+CN^2} = \blue {\frac 8{\sqrt 5}}$ .

The area of $\triangle ABC$ is given by:

$\begin{aligned} [ABC] & = \frac 12 \cdot CA \cdot BC \cdot \sin \angle ACB \\ & = \frac 12 \cdot 2 \cdot CP \cdot \cos \angle PCA \cdot BC \cdot \sin \angle ACB \\ & = CP \cdot BC \cdot \cos (\angle PCT - \angle ACT) \sin (\angle ACT + BCT) & \small \blue{\text{Let }\angle ACT = \theta} \\ & = \blue{\frac {32}{\sqrt 5} \cos \left(\tan^{-1} \frac 34 - \theta \right) \sin \left(\theta + \tan^{-1} \frac 12 \right)} \\ & = \frac {32}{\sqrt 5} \left(\frac 35 \sin \theta + \frac 45 \cos \theta \right) \left(\frac 2{\sqrt 5} \sin \theta + \frac 1{\sqrt 5} \cos \theta \right) \\ & = \frac {32}{25} \left(3\sin \theta + 4\cos \theta \right) \left(2 \sin \theta + \cos \theta \right) \\ & = \frac {32}{25} (6\sin^2 \theta + 11 \sin \theta \cos \theta + 4 \cos^2 \theta) \end{aligned}$

Similarly,

$\begin{aligned} [CDE] & = \blue{\frac 4{\sqrt 5} \cos \left(\tan^{-1} \frac 34 - \theta \right) \sin \left(\tan^{-1} 2 - \theta \right)} \\ & = \frac 4{25} (-3\sin^2 \theta + 2\sin \theta \cos \theta + 8 \cos \theta) \end{aligned}$

Then the sum of areas of the two triangles:

$\begin{aligned} S & = [ABC]+[BCD] \\ & = \frac 45 (9\sin^2 \theta + 18 \sin \theta \cos \theta + 8 \cos^2 \theta) \\ & = \frac 25 (18 \sin^2 \theta + 36 \sin \theta \cos \theta + 16 \cos^2 \theta ) \\ & = \frac 25 (\sin^2 \theta - cos^2 \theta + 36 \sin \theta \cos \theta + 17) \\ & = \frac 25 (18 \sin 2\theta - \cos 2\theta + 17) \\ & = \frac 25 \left(5\sqrt{13} \cos \left(2\theta - \tan^{-1} \frac 1{18} \right) + 17\right) \end{aligned}$

Therefore $S$ is maximum when $2\theta - \tan^{-1} \dfrac 1{18} = \dfrac \pi 2$ $\implies \theta = \dfrac \pi 4+ \dfrac 12 \tan^{-1} \dfrac 1{18}$ . Then the ratio of

$\begin{aligned} \frac {[CDE]}{[ABC]} & = \frac \blue{\frac 4{\sqrt 5} \cos \left(\tan^{-1} \frac 34 - \theta \right) \sin \left(\tan^{-1} 2 - \theta \right)}\blue{\frac {32}{\sqrt 5} \cos \left(\tan^{-1} \frac 34 - \theta \right) \sin \left(\theta + \tan^{-1} \frac 12 \right)} \\ & = \frac {\sin \left(\tan^{-1} 2 - \theta \right)}{8\sin \left(\theta + \tan^{-1} \frac 12 \right)} \\ & = \frac {\sqrt{\frac 1{26}(13-3\sqrt{13}}}{8\sqrt{\frac 1{26}(13+3\sqrt{13}}} = \frac {13-3\sqrt{13}}{16\sqrt{13}} = \frac {\sqrt{13}-3}{16} \end{aligned}$

Therefore $\sqrt[5]{p+q+m} = \sqrt[5]{13+3+16} = \boxed 2$ .

We start by positioning the circles. The smaller is tangent to the y axis with center at (1,1) and the bigger circle is tangent to the smaller with center at (5,4)

Equating $(x-1)^{2}+(y-1)^{2}=1$ and $(x-5)^{2}+(y-4)^{2}=16$ , gives point $(\frac{9}{5}, \frac{8}{5})$ as intersection.

Now we have 2 fixed points for each triangle, the intersection point and the points where the circles are tangent to the x axis, (1,0) and (5,0).

Looking at the bigger circle, the third point needed to use the area formula is just $(x, f(x))$ . We use the top semi-circle because it will create a bigger triangle, so $f(x)=\sqrt{-(x-9)(x-1)}+4$ . For the smaller circle, we need a point that can be mapped from the other that is traveling around the bigger circle.

$x=-4x_{2}+9 ⟹ x_{2}=\frac{9-x}{4}$

We insert this modified $x$ into the bottom semi-circle function to get point $(x_{2}, f(x_{2}))$ with $f(x_{2})=1 - \frac{\sqrt{(1 - x)(x - 9)}}{4}$

After plugging these points into the formula provided at the top, we get the equations for the area of both triangles as a function of $x$ . Long story short: