Embiggen this Circle

Let us plot this into a Cartesian plane, with $A(0,0), B(w,0), C(w,1), D(1,0), E(w, \tfrac12), F(\tfrac w2, 0)$ .

The coordinate of the $G$ can be found by finding the intersection point between the two straights $AE$ and $DF$ :

$\begin{cases} \text{Straight line }AE: & \quad y = \frac12 \cdot \frac xw = \frac x{2w} \\ \text{Straight line }DG: & \quad \frac x{w/2} + \frac y1 = 1 \end{cases} \quad \Rightarrow \quad G = \left( \frac {2w}5, \frac15 \right)$

Let $r$ denote the radius of the circle, then $r = \dfrac{ \mathcal A}{s}$ , where $\mathcal A$ and $s$ denotes the area and semiperimeter of the triangle $DEG$ , respectively.

$\mathcal A$ can be computed by finding the complementary area. That is, $(\text{Area of rectangle } ABCD) - \mathcal A = (\text{Area of triangle } DAF) + (\text{Area of triangle } ABF) + (\text{Area of triangle } DCE) - (\text{Area of triangle } GAF)$

In the right-hand-side of the equation above, the first three areas are easy to compute because they are all right-angle triangles. The last area can be computed using shoelace formula .

Simplify all these, we get $\mathcal A = \dfrac{3w}{10} \tag1$

By Pythagorean theorem , we can find all the side lengths of the triangle $DEG$ . Thus, we can obtain $s = \dfrac{DE + EG + GD}2 = \dfrac15\left (\sqrt{4 + w^2 } + 2 \sqrt{1+4w^2} \right ) \tag2$

Let $P$ denote the ratio of the area of the circle to the area of the triangle, so $P = \dfrac{\pi r^2}{1\cdot w} = \dfrac94 \cdot \dfrac w{\left(\sqrt{4 + w^2 } + 2 \sqrt{1+4w^2} \right)^2 }$

Maximizing $P$ is equivalent to minimizing $Q := \dfrac {\left(\sqrt{4 + w^2 } + 2 \sqrt{1+4w^2} \right)^2 } w = 17w + \dfrac{4\sqrt{w^2 + 4} \cdot \sqrt{4w^2 + 1}}w + \dfrac8w$

At its critical point(s), $\dfrac{dQ}{dw} = 0$ . Apply quotient rule and simplify all the tedious algebra, we get $w = \sqrt{ \dfrac{\sqrt{481} - 17}8 }$ .

Obviously $Q$ has no maximum value because we can set $w$ to be unboundedly large, so the critical point above must be a minimum value.

Hence, the ratio in question is maximized when $w = \sqrt{ \dfrac{\sqrt{481} - 17}8 }$ . The answer is $481 + 17 + 8 = \boxed{506} .$

Let's use sympy to find the solution of $w = \sqrt \dfrac{-17 + \sqrt {481}}{8} \implies 17 + 481 + 8 = \boxed{ 506}$

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# https://www.geogebra.org/classic/crs42wcz
# https://brilliant.org/problems/embiggen-this-circle/

from sympy import *

width = symbols('width')
height = 1

A = Point(0, 0)
B = Point(width, 0)
C = Point(width, height)
D = Point(0, height)
F = Point(width/2, 0)           # midpoint AB
E = Point(width, height/2)      # midpoint BC
ae = Line(A, E)
df = Line(D, F)
G = ae.intersection(df)[0]
radius = Triangle(G, E, D).inradius

# we wish to maximize the ratio of the areas
ratio = radius*radius / (width*height)

# differentiate w.r.t width
dif = diff(ratio, width)

# check the 2nd derivative
dif2 = diff(dif, width)

# solve for extrema
sol = solve(dif, width)
for i in sol:
    if i > 0:
        print("width =", i)
        print("2nd derivative < 0?", dif2.subs(width, i) < 0)

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# width = sqrt(-17/8 + sqrt(481)/8)
# 2nd derivative < 0? True