Dynamic Geometry: P119

From the compiled calculations in Dynamic Geometry: P96 Series , we know the radius of the circle is $6.5$ , and the dividing chord $AB=12$ . If the center $O$ of the center is the origin $(0,0)$ of the $xy$ -plane, then $A=(-6, 2.5)$ and $B = (6,2.5)$ . Let the upper and lower diametrically opposite moving points be $C$ and $C'$ . If $CC'$ makes an angle of $\theta$ with the $x$ -axis, then $C(6.5 \cos \theta, 6.5 \sin \theta$ and the height of the upper triangle $h = 6.5 \sin \theta - 2.5$ . Similarly, the height of the lower triangle $h' = 6.5 \sin \theta + 2.5$ . From the compiled calculations, the side lengths of the upper and lower squares are:

$\begin{cases} s = \dfrac {12h}{12+h} = \dfrac {156\sin \theta - 60}{13\sin \theta +19} \\ s' = \dfrac {12h'}{12+h'} = \dfrac {156\sin \theta + 60}{13\sin \theta +29} \end{cases}$

When $\dfrac s{s'} = \dfrac {2077}{4077}$ . we have:

$\begin{aligned} \frac {156\sin \theta - 60}{13\sin \theta +19} \cdot \frac {13\sin \theta +29}{156\sin \theta + 60} & = \frac {2077}{4077} \\ 4077(2028\sin^2\theta+3744\sin \theta -1740) & = 2077(2028\sin^2\theta+3744\sin \theta +1140) \\ 4077(169\sin^2\theta+312\sin \theta -145) & = 2077(169\sin^2\theta+312\sin \theta +95) \\ 4225\sin^2 \theta +7800\sin \theta-9856 & = 0 \\ (65 \sin \theta - 56) (65 \sin \theta + 176) & = 0 \qquad \qquad \small \blue{\text{Since }\sin \theta > 0} \\ \implies \sin \theta & = \frac {56}{65} \end{aligned}$

The ratio of the areas of the two triangles

$\frac {A_\triangle}{A'_\triangle} = \frac {\frac 12 \cdot 12h}{\frac 12 \cdot 12h'} = \frac h{h'} = \frac {6.5\cdot \frac{56}{65}-2.5} {6.5\cdot \frac{56}{65}+2.5} = \frac {5.6-2.5}{5.6+2.5} = \frac {3.1}{8.1} = \frac {31}{81}$

Therefore $p+q = 31+81 = \boxed{112}$ .

The total height of the circle equals 13, so its radius $r=\frac{13}{2}=6.5$ .

Let's set the centre of the circle at the origin, then

• The circle has equation $x^2+y^2=\frac{169}{4}$ .
• The yellow line has equation $y=\frac{5}{2}$ .

The yellow line intersects the circle at $x=\pm 6$ , so its length equals 12.

Now if we fit a rectangle of height y inside a triangle of height h and width w, the width of the triangle can be $x=(h-y)\frac{w}{h}$ In the special case that the rectangle is a square (with side $s=x=y$ ), this simplifies to $s=\frac{wh}{w+h}$

Now for both triangles $w=12$ and we know that $h_{green}=h_{blue}+5$

Setting $h_{blue}=h$ and $h_{green} =h+5$ , we can now set the given ratio $\frac{4077}{2077}=\frac{(h+5)(12+h)}{(17+h)h}$ Which is in effect a quadratic in h with positive solution $h=\frac{31}{10}$

The triangle area ratio now is just the height ratio $\frac{h}{h+5}=\frac{31}{81}$

The numbers 31 and 81 are coprime and their sum is $\boxed{112}$