Four Green Two Blue Circles

Let $A$ , $G$ , and $J$ be centers of green circles with radii $r_1$ , $C$ be the center of a blue circle with radius $r_2$ , and $O$ be the center of the large black circle with radius $R$ . Let $BC$ be an altitude of $\triangle ACO$ and $FG$ be an altitude of $\triangle GHO$ , both perpendicular to $AH$ . Finally, let $D$ be the point of tangency of circle $A$ and the red line, $E$ be the point of tangency of circle $C$ and the red line, $E'$ be the point of tangency of circle $G$ and the red line, and $H$ be the point of intersection of circle $J$ and $AJ$ , as pictured below.

Let $OD = t$ . By symmetry, $OH = t$ . Then $OJ = OH + HJ = t + r_1$ , and since $OG = R - r_1$ and $OJ = R - r_1$ , $OG = OJ = t + r_1$ . Also, $DJ = OD + OH + HJ = 2t + r_1$ , and since $DF = E’G = r_1$ , $FJ = DJ - DF = 2t + r_1 - r_1 = 2t$ and $OF = DF - OD = r_1 - t$ . In addition, $GJ = 2r_1$ , and let $FG = s$ .

Then using Pythagorean’s Theorem on $\triangle OFG$ and $\triangle GFJ$ , we have $FG^2 + OF^2 = OG^2$ and $FG^2 + FJ^2 = GJ^2$ , or:

$s^2 + (r_1 - t)^2 = (r_1 + t)^2$

$s^2 + (2t)^2 = (2r_1)^2$

which solves to $s = \sqrt{2\sqrt{5} - 2}r_1$ and $t = \frac{1}{2}(\sqrt{5} - 1)r_1$ .

Since $BO = OD + BD$ , and $BD = CE = r_2$ , $BO = t + r_2$ and $AB = AD - BD = r_1 - r_2$ . Also, $OC = R - r_2$ and $AC = r_1 + r_2$ . Since $OJ = R - r_1 = t + r_1, R = t + 2r_1$ , so $OC = t + 2r_1 - r_2$ . Let $BC = u$ .

Then using Pythagorean’s Theorem on $\triangle ABC$ and $\triangle OBC$ , we have $AB^2 + BC^2 = AC^2$ and $BO^2 + BC^2 = OC^2$ , or:

$(r_1 - r_2)^2 + u^2 = (r_1 + r_2)^2$

$(t + r_2)^2 + u^2 = (t + 2r_1 - r_2)^2$

which with the above $t$ value of $t = \frac{1}{2}(\sqrt{5} - 1)r_1$ solves to $u = \sqrt{2\sqrt{5} - 2}r_1$ and $r_2 = \frac{1}{2}(\sqrt{5} - 1)r_1$ (which makes $s = u$ and $t = r_2$ ).

Therefore, $\frac{r_1}{r_2} = \frac{r_1}{\frac{1}{2}(\sqrt{5} - 1)r_1} = \frac{1 + \sqrt{5}}{2} = \phi$ , the golden ratio. This means that $a = b = 1$ , $d = 5$ , and $c = 2$ , and $a \times b \times c \times d = \boxed{10}$ .

The ratio is the Golden Ratio $\dfrac{1+\sqrt{5}}{2}$ . See Vreken's excellent solution. Below is how it would look like if the plane was filled with the same green and blue circles, with one red tangent line and purple tangent circle added.

From this we can set up the following equation and solve for $r_1$ in terms of $r_2$

$\sqrt{ (r_1+r_2)^2 - (r_1-r_2)^2 + (2r_2)^2 }=2r_1$

and thus get us the ratio

$r_1=(\dfrac{1+\sqrt{5}}{2})r_2$

but this seemingly quick solution makes a number of improper assumptions, even if all true. Vreken's solution avoids making those unwarranted assumptions, such as noting that tangent points $E, E'$ are not necessarily the same point. The other approach would be to show that it is possible to assemble in this manner green and blue circles in that ratio, and then show that a [purple] circle centered at any blue circle would be tangent to four green and two blue circles, red chord included.

Let the center of the black circle be $O$ and its radius be 1, centers of three green circles be $A$ , $C$ and $D$ and that of a blue circle be $B$ . We note that $A$ , $C$ and $D$ are equidistance from $O$ , therefore $\triangle ACD$ is inscribed in a circle and $CD$ is a diameter and hence $\angle CAD = 90^\circ$ . Also note that $AB || CD$ . Let $\angle ADC = \theta$ ; then $\angle CAB = \theta$ . Since $AB=BC= r_1+r_2$ , $\triangle ABC$ is isosceles and $\angle BCA = \theta$ . $\triangle BCF$ is a reflection of $\triangle OCF$ , therefore, $CB=CO$ $\implies r_1+r_2 = 1-r_1$ , $\implies \color{#3D99F6} r_2 = 1-2r_1$ .

Now $\sin \theta = \dfrac {AD}{CD} = \dfrac {2r_1}{2(1-r_1)} = \dfrac {r_1}{1-r_1}$ . Also $\sin \theta = \dfrac {DE}{AD} = \dfrac {CD-\color{#D61F06}CE}{AD} = \dfrac {2-2r_1 - \color{#D61F06}2r_1}{2r_1} = \dfrac {1-2r_1}{r_1}$ . Note that $\color{#D61F06} CE$ is the sum of $\color{#D61F06}2r_1$ of green circles centered $A$ and $C$ .

Then we have:

$\begin{aligned} \frac {r_1}{1-r_1} & = \frac {1-2r_1}{r_1} \\ r_1^2 & = 2r_1^2 - 3r_1 + 1 \\ r_1^2 - 3r_1 + 1 & = 0 \\ \implies r_1 & = \frac {3-\sqrt 5}2 & \small \color{#3D99F6} \text{Note that }r_1 < 1 \\ r_2 & = 1 - 2r_1 = \sqrt 5 - 2 \end{aligned}$

$\begin{aligned} \implies \frac {r_1}{r_2} & = \frac {3-\sqrt 5}{2(\sqrt 5-2)} = \frac {(3-\sqrt 5)(\sqrt 5+2)}2 = \frac {1+\sqrt 5}2 = \varphi \end{aligned}$

Therefore, $a\times b \times c \times d = 1\times 1 \times 2 \times 5 = \boxed{10}$ .

Set the origin at the center of the black circle (having radius R), and scale the axes such that the green circles have their centers at distance 1 from the origin. Three of these centers are A=(0,1), B=(0,-1) and C=(x,y). The green circles with centers B and C are touching, so that $(2r_1)^2=BC^2=x^2+(y - -1)^2=x^2+y^2+2y+1=2+2y$ (using the fact that $1=OC^2=x^2+y^2$ ), so that $y+1=2r_1^2$ We also have the horizontal red line touching both green circles so that $y+r1=1-r1$ or $y=1-2r_1$ . Substitute this above and we get $r_1^2+r_1-1=0$ which solves as $r_1=\frac{\sqrt{5}-1}{2}$ Note that $r_1=\phi-1=\frac{1}{\phi}$ . We now also know $y=2-\sqrt{5}$ and $R=1+r_1=\phi$ . Next, let's set D=(a,b) as the center of the blue circle, (touching the red line and top green circle). We have $AD=r_1+r_2$ , $OD+r_2=R$ and $b=1-r_1+r_2$ , from which we solve as follows: $(r_1+r_2)^2=AD^2=a^2+(1-b)^2=a^2+b^2-2b+1$ , $(R-r_2)^2=OD^2=a^2+b^2$ . Subtracting these gives $(r_1+r_2)^2-(R-r_2)^2=-2b+1$ ,

$r_1^2+2r_1r_2-R^2+2Rr_2=-2b+1=-2+2r_1-2r_2$

We already know $r_1$ and $R$ , filling these values in and working out gives

$r_2=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{2r_1}{\sqrt{5}+1}$

So that ${\frac{r_1}{r_2}=\phi=\frac{1+\sqrt{5}}{2}}$

The requested answer is 1×1×2×5=10.

Note that $...r_2,r_1,1,R,...$ are part of a geometric series with factor $\phi$ .

Finally, by calculating x and a, we can now verify that the circles with centers C and D are not only touching the red line, but also touching each other, something I did not presume. They indeed do: $a=x=\sqrt{4\sqrt{5}-8}$ .

One can show, R=2(r1)+r2 and ratio of (r1/r2) = x, satisfies

x^2-x-1=0, root x =(1/2)(1+√5),

Answer=10

According to the figure, let:

• $A$ , $B$ , $C$ , and $D$ be the centers of green circles
• $E$ be the center of one of the blue circles
• $O$ be the center of the big circle, and its radius be $R$
• $P$ be the point where the green circle centered at $A$ touches the chord
• $Q$ be the point where both the blue circle and a green circle touch each other and the chord
• $M$ be the intersection of lines $AD$ and $PQ$ .

Lines $AP$ and $DQ$ are both perpendicular to the chord so $AP || DQ$ and because $AP = DQ = r_1$ , $APDQ$ is a parallelogram. Point $M$ is the intersection of diagonals in $APDQ$ , so $PM = QM$ and $AM = DM$ (I assumed that $P$ and $Q$ are not the same points, because of the figure given in the problem).

The distance from $O$ to each of the points $A$ , $B$ , $C$ , and $D$ is $R - r_1$ , so these four points lie on the same circle. Because of symmetry we have that $AC$ is the diameter of this circle and with that, $\angle ABC = \angle ADC = 90°$ . Line $OM$ connects the midpoints of $AC$ and $AD$ , so $OM || CD$ and $OM =\dfrac{1}{2}CD = \dfrac{1}{2}2r_1 = r_1$ .

From $OM || CD$ we have $\angle AMO = \angle ADC = \angle APM = 90°$ and with $\angle AOM = \angle MOP$ , we have $\triangle AMO \sim \triangle APM$ and so: $\dfrac{AP}{PM} = \dfrac{AM}{OM}$ . Let $PM = QM = x$ . We know that:

$\begin{aligned} AP & = r_1 \\ PM & = x \\ AM & = \sqrt{AP^2 + PM^2} = \sqrt{r_1^2 + x^2} \\ OM & = r_1 \\ \dfrac{AP}{PM} & = \dfrac{AM}{OM} \\ \dfrac{r_1}{x} & = \dfrac{\sqrt{r_1^2 + x^2}}{r_1} \\ \dfrac{r_1^2}{x^2} & = \dfrac{r_1^2 + x^2}{r_1^2}\end{aligned}$

Here, we have a very famous equation which tells us that the ratio of $r_1^2$ and $x^2$ might exactly be the golden ratio (or golden number, $\phi = \dfrac{1 + \sqrt{5}}{2}$ ), but for that to be true we first need to prove that $r_1^2 > x^2$ . This can be proved if we take a look at one specific triangle:

Let $H$ be a point on $AP$ so that $EH \perp AP$ . We have:

$\begin{aligned} EH & = PQ = PM + MQ = 2x \\ PH & = EQ = r_2 \\ AH & = AP - PH = r_1 - r_2 \\ AE & = r_1 + r_2 \end{aligned}$

And by triangle inequality:

$\begin{aligned} EH & < AH + AE \\ 2x & < r_1 - r_2 + r_1 + r_2 \\ 2x & < 2r_1 \\ r_1 & > x \\ r_1^2 & > x^2 \end{aligned}$

And thus: $\boxed{ \dfrac{r_1^2}{x^2} = \phi \Rightarrow x = \dfrac{r_1}{\sqrt{ \phi}} }$

Now, we know that $OA = OD = R - r_1$ , $AE = ED = r_1 + r_2$ so $\triangle OAE \cong \triangle ODE$ from which we deduce that $OA || DE$ and $AE || OD$ . This gives us: $OA = AE = DE = OD$ i.e. $OAED$ is a rhombus with $M$ being the intersection of its diagonals as it is the midpoint of $A$ and $D$ and $\angle AME = 90°$ since diagonals in a rhombus are perpendicular to each other. This leads us to $\angle EMQ = 180° - \angle AME - \angle AMP = 90° - \angle AMP = \angle MAP$ and we have another pair of similar triangles: $\triangle APM \sim \triangle MQE$ . Finally:

$\begin{aligned} \dfrac{AP}{MP} & = \dfrac{MQ}{EQ} \\ \dfrac{r_1}{x} & = \dfrac{x}{r_2} \\ r_1 r_2 & = x^2 \\ r_1 r_2 & = \dfrac{r_1^2}{\phi} \\ r_2 & = \dfrac{r_1}{\phi} \\ \end{aligned}$ $\boxed{ \dfrac{r_1}{r_2} = \phi = \dfrac{1 + \sqrt{5}}{2}}$