Another hard geometry puzzle
In Figure 1, an infinite number of blue circles are drawn in the region bounded by the two semicircles and quarter-circle. The square, as shown in Figure 1, has unit length.
The solution to the puzzle is a general formula for the radius of the nᵗʰ largest circle in terms of n.
At first glance, this question looked too hard for me, and while I managed to solve all the other questions in the set that this question came from, I was too afraid to even get started on this one because of the tediousness that I anticipated I would have to go through to get an answer. But for you, dear reader, I will go through this anticipated tediousness so that you may have the experience of reading an account of my frustrations in going through the experience of coming up with a solution to this puzzle. Rest assured, dear reader, that I have not consulted any source when trying to find a solution, so all the frustrations I express in this article are genuine. My hope is that the genuineness of my frustrations will increase your enjoyment and thus provide further value to you.
We will be using a coordinate system to make keeping track of where everything is in relation to everything else easier. The origin of this coordinate system is located in the bottom-left corner of the unit square, the rightward direction is that of the positive x-axis and the upward one is that of the positive y-axis.
With the coordinate system set up, the next thing we must deal with is the first circle, the one with radius r(1). But before that, let me write a few lines on the notation I use in this article to denote the radius of the nᵗʰ largest circle. According to my notation, r denotes a sequence whose nᵗʰ term is r(n). A sequence is actually a function that takes the term number, a positive integer, as its argument. Thus, the domain of the function r is the set of all positive integers (Z₊) while its range is the set of all real numbers (R).
That bit of technicality was for all the math pedants out there. With that out of the way, we may now continue the torture of my soul with this devilish puzzle.
Where is the centre of the circle with radius r(1)?
This is the question we will try answering first. Now, there are three variables that completely specify a circle, the x-coordinate of its centre, the y-coordinate of its centre and its radius. So, we expect there to be three independent equations if we have any hope of finding the location of the circle with radius r(1). And, lucky for us, indeed there are three.
The three red line segments you see in Figure 2 are the bases of the three equations that will help us completely solve for the position (i.e, the coordinates of the centre) and size (i.e., radius) of the circle with radius r(1).
One point we must touch on before, is why the line segment remains straight when crossing the boundary between touching circles. The reason is that the two circles share a tangent at that point, and that tangent is perpendicular to both the circle’s radii. So the two 90° angles formed by the two radii and common tangent form a linear pair (i.e., the sum of the angles is 180°), and thus the radii are a part of the same straight line.
Why is the tangent to a circle perpendicular to the circle’s radius?
To explain why a tangent is perpendicular to the circle’s radius, we’d have to go on a little detour. After we’re done with that detour, we’ll come back to this puzzle.
We suppose, for the sake of supposition, that the radius to the circle is not perpendicular to the tangent line. So, as Figure 3 shows, the radius of the circle is supposed to be OT, while a perpendicular called OP is dropped from O to the tangent line at P.
Now, it is a well-known fact that the shortest line segment from a point to a line is one that is perpendicular to the line.
Why is the perpendicular line segment from a point to a line the shortest one to the line?
This can be proved from Figure 3 by imagining there to be no circle around the point O. In the triangle OTP, we have supposed that the perpendicular OP is not the shortest. But any other line that can be drawn from point O to the line below it will form a triangle of the form OTP where OT > OP simply because OT is the hypotenuse (which is the longest side of a right-angled triangle because it is the square root of the sum of the squared arm lengths of said triangle). Thus, either all the lines from O to the line are of the same length (which is not true), or OP is the shortest (which is true).
So now, we know that OP in Figure 3 is the shortest line segment from the centre to the tangent line because it is perpendicular. Since the tangent line touches the circle at only one point, and the perpendicular OP touches the tangent line, the perpendicular OP must be either of radial length or longer. If OP were shorter, it would remain inside the circle while the tangent line would be outside the circle (or on the circle for that one point where it touches the circle), so it would be impossible for them to meet.
But OT was the radius of the circle, so OP is of radial length or longer and OT is of radial length. Thus, either OP = OT or OP > OT. If they’re equal, then OTP forms an isosceles triangle with OT = OP. OT cannot be smaller than OP without T and P coinciding because OT is not the shortest line segment from point O to line TP.
There is a theorem according to which angle OTP = angle OPT if OP = OT. Let us also prove that angle OTP > angle OPT if OT > OP even though it isn’t required for our purposes because I wrote it up before I realised I didn’t actually need it and I can’t complain about the minutes this will add to this article’s estimated reading time.
Why are the angles opposite to equal sides in a triangle equal?
To prove this theorem, we draw an isosceles triangle as shown in Figure 4 and construct a line segment from the vertex shared by the two equal sides to the base opposite said vertex. This line segment is drawn such that it divides TP into two equal halves. Such line segments are called medians of a triangle, and there are three possible medians that can be drawn (one from each vertex).
Now, the triangles OTA and OPA are congruent because of the SSS congruency criterion. The SSS congruency criterion is true because only one triangle can be drawn when its three sides are specified. To draw such a triangle, one chooses one of the sides and draws it as the base of the triangle. Then, one uses a compass to draw an arc of the length of the second side from one of the ends of the chosen base, and an arc of the length of the third side from the other end of the base. If the lengths are chosen so that they follow the triangle inequality (the sum of any two of them is greater than the third one), the arcs will intersect at one point on one side of the base at at another point on another side of the base. Those points represent the possible positions of the third point, and the two triangles so formed are just mirror images of each other. So, in effect, only one triangle can be drawn with three specified lengths, so all triangles with those three specified lengths must be the same as each other in shape and size. They must be congruent.
Because OTA and OPA are congruent, angle OTP = angle OPT.
If the triangle is not isosceles and OT > OP, then we can use the above theorem to conclude that angle OPT > angle OTP.
Why is the angle opposite to the greater side in a triangle greater than the angle opposite to the smaller side?
I adapted the following proof from youtu.be/LeeiVVAoPUk because I found some problems with the one I was originally going to present.
In Figure 5, the case of OT > OP is presented. We have to prove that angle OPT > angle OTP. A construction of line segment PA is made so that OA = OP. The first fact is that angle OAP = angle OPA because OA = OP, from the theorem we just proved above. The second fact is that angle OTP + angle APT = angle OAP. This is because the sum of the three angles of triangle ATP are 180° and the angles TAP and OAP are supplementary because they form a linear pair.
Now, we will prove that the sum of the three angles of a triangle is 180°.
Why is the sum of the three interior angles of a triangle 180°?
In Figure 6 above, ED and BC are parallel lines. With AB as a transversal, angle ABC = angle BAE because they are alternate interior angles. Similarly, with AC as a transversal, angle ACB = angle CAD because they are, once again, alternate interior angles.
Since, the sum of the angles EAB, BAC and CAD is 180° (because they form an angle from one end of the line to the other), the sum of the three angles of triangle ABC (namely ABC, BAC and ACB) is 180° as well.
So, going back to Figure 5 (which I’ve shown here again), we see that angle OTP + angle APT = angle OAP, but angle OAP = angle OPA because OP = OA by construction. Thus, angle OTP + angle APT = angle OPA.
We can add angle APT to both sides of this equality without affecting it, so we do it to get the following equation:
angle OTP + 2(angle APT) = angle OPA + angle APT
Since angle OPA + angle APT = angle OPT, we can rewrite the above equality as following one:
angle OTP + 2(angle APT) = angle OPT
Since OT > OP, we know that OT > OA (since OA = OP by construction). Thus, A must lie on line segment OT, which means angle APT is a positive angle. The above equality, therefore, proves that angle OPT > angle OTP.
Wow, what a journey. We can finally go back to Figure 3 (which I’ve shown here again) and say that OT = OP (we proved this earlier in this article). Thus, angle OTP = angle OPT, but angle OPT = 90°, but if angle OTP = 90° as well, then the angle sum property for triangle OTP will be violated for any non-zero value for angle TOP.
If angle TOP = 0°, then triangle OTP is degenerate and T coincides with P, which means the radius is perpendicular to the tangent of the circle.
Now, with all the necessary (and unnecessary) justifications made, we can go back to our beloved puzzle from Figure 2.
Using Figure 2’s red line segments from the centres of the two semicircles and one quarter-circle to either the centre of the circle with radius r(1) or through the same point to the point of tangency, we can come up with three independent equations. In the below equations, (x,y) is the coordinate of the centre of the circle with radius r(1).
(0.5 + r(1))^2 = (0.5 — x)^2 + y^2 [Equation 1]
(0.5-r(1))^2 = x^2 + (y-0.5)^2 [Equation 2]
(1-r(1))^2 = (1-x)^2 + y^2 [Equation 3]
I won’t go through the trouble of solving these equations here… Oh who am I kidding, I’ll solve them right here.
2(r(1)) = 0.5(0.5-2x) + 0.5(2y-0.5) [Equation 1-2]
(2(r(1))-0.5)(1.5) = (-0.5)(1.5-2x) [Equation 1-3]
After one round of simplifying, we get the following two equations:
2(r(1)) = y-x [Equation 1-2]
x = 3(r(1)) [Equation 1-3]
Simplifying [Equation 3] yields the following:
1 - 2(r(1)) + (r(1))^2 = 1 — 2x + x^2 + y^2 [Equation 3]
Substituting [Equation 1-3] into [Equation 3] yields the following:
4(r(1)) — 8(r(1))^2 = y^2 [Equation 3<(1-3)]
At this stage, we can already say that r(1) < 1 because if it wasn’t, y wouldn’t be a real number according to [Equation 3<(1-3)].
Substituting [Equation 3<(1-3)] and [Equation 1–3] into [Equation 1-2] yields the following:
25(r(1))^2 = 4(r(1)) — 8(r(1))^2
Since r(1) > 0, we can divide the equation by r(1) to get the following:
r(1) = 4/33
What is different about the smaller circles (with radii r(2), r(3), r(4), etc. from the largest one (with radius r(1))?
Now, to simplify calculating r(2) and eventually r(n), we need to see which equations that applied to the circle with radius r(1) also apply to the circle with radius r(2), and it immediately stands out to us that the only difference between the circle with radius r(1) and the rest of the circles is [Equation 2].
Figure 7 shows this fact in a very succinct manner using blue line segments. [Equation 2] is replaced by a relation between r(2) and r(1) and, in the general case, a recurrence relation between r(n) and r(n-1). I use x(n) and y(n) below to denote the x- and y-coordinate of the centre of the circle with radius r(n).
Let’s call this new equation [Equation 4]:
(r(1) + r(2))^2 = (x(1) — x(2))^2 + (y(1) — y(2))^2 [Equation 4]
The other two equations are as follows (in heavily simplified form):
4(r(2)) — 8(r(2))^2 = (y(2))^2 [Equation 5]
x(2) = 3(r(2)) [Equation 6]
Substituting [Equation 5] and [Equation 6] into [Equation 4] yields the following:
(r(1) + r(2))^2 — (3*r(1) — 3*r(2))^2 = (sqrt(4*r(1) — 8*r(1)^2) — sqrt(4*r(2) — 8*r(2)^2))^2
One round of heavy simplification yields the following:
r(1) + r(2) — 5*r(1)*r(2) = 2*sqrt((r(1) — 2*r(1)^2) * (r(2) — 2*r(2)^2))
Squaring both sides yields the following:
r(1)^2 + r(2)^2 + 25 * r(1)^2 * r(2)^2 + 2*r(1)*r(2) — 10*r(1)*r(2)*(r(1)+r(2)) = 4*r(1)*r(2) — 8*r(1)*r(2)*(r(1)+r(2)) + 16 * r(1)^2 * r(2)^2
Another round of simplification yields the following:
r(1)^2 + r(2)^2 + 9 * r(1)^2 * r(2)^2 — 2*r(1)*r(2) — 2*r(1)*r(2)*(r(1)+r(2)) = 0
Rearranging the terms a little but yields the following:
(9*r(1)^2 — 2*r(1) + 1)*r(2)^2 — (2*r(1) + 2*r(1)^2)*r(2) + r(1)^2 = 0
We can solve for r(2) using the quadratic formula.
But first, we must derive it. Given a general quadratic equation of the form ax^2 + bx + c = 0, we try to “complete the square” as follows:
x^2 + 2 * (b/2a) * x + (b/2a)^2 + c/a = (b/2a)^2
(x + b/2a)^2 = (b^2— 4ac)/(4a^2)
x + b/2a = ± sqrt(b^2-4ac)/2a
x = (-b ± sqrt(b^2-4ac))/2a
So, we were originally looking at the following equation:
(9*r(1)^2 — 2*r(1) + 1)*r(2)^2 — (2*r(1) + 2*r(1)^2)*r(2) + r(1)^2 = 0
The above equation is quadratic in r(2). Solving it for r(2) using the quadratic formula yields the following recurrence relation:
r(2) = (r(1) + r(1)^2 ± 2*r(1)*sqrt(r(1) — 2*r(1)^2)) / (9*r(1)^2 — 2*r(1) + 1)
Well, we’re stuck here, so we go back to a previous equation and pull a rabbit out of a hat to get this puzzle solved as soon as possible because I’m so tired that I need to go to bed soon or I’ll literally drop to the floor , asleep.
r(1)^2 + r(2)^2 + 9 * r(1)^2 * r(2)^2 — 2*r(1)*r(2) — 2*r(1)*r(2)*(r(1)+r(2)) = 0
The God of intuition has decided to grant me a favour because of the strength of my fire of desire to solve this puzzle as soon as possible so I can go to bed and sleep. This God has commanded me to divide throughout by r(1)^2 * r(2)^2 and I couldn’t explain why even if I wanted to (ask the God of intuition for help). This operation yields the following:
1/r(1)^2 + 1/r(2)^2 + 9 — 2/(r(1) * r(2)) — 2*(1/r(1) + 1/r(2)) = 0
Rearranging the terms a little yields the following:
1/r(2)^2 - (2/r(1) + 2)*(1/r(2)) + 1/r(1)^2 — 2/r(1) + 9 = 0
This is a quadratic equation in 1/r(2), and using the quadratic formula one obtains the following:
1/r(2) = 1/r(1) + 1 ± 2*sqrt(1/r(1) - 2)
So, 1/r(2) = 57/4 or 1/r(2) = 17/4. Since, the radii are supposed to get smaller because n represents the circle with the nᵗʰ largest radius, the multiplicative inverse of the radii should get larger so 57/4 is the right choice. In other words, the following equation is correct:
1/r(2) = 1/r(1) + 1 + 2*sqrt(1/r(1) - 2)
There is a rule for recurrence relations that states that the power of the closed form of the recurrence is one greater than its power in the recurrence relation. I’m sure there’s a reason and explanation behind this rule, but it is utterly beyond me to find that out now. I’m already more than 4000 words in, and I really feel like the God of sleep is pulling my eyelids shut. I’m fighting to keep them open. That should be more than enough evidence that I’m really tired, so please cut me some slack!
Since the recurrence relation’s highest power is linear, the closed form (if it exists), is quadratic!
1/r(n) = an^2 + bn + c
The following values of r(n) have been calculated from the recurrence relation established the previous paragraph or a couple of ones before that.
1/r(1) = 33/4, so 33/4 = a + b + c [Equation 7]
1/r(2) = 57/4, so 57/4 = 4a + 2b + c [Equation 8]
1/r(3) = 89/4, so 89/4 = 9a + 3b + c [Equation 9]
Using the three equations above to get [Equation 9-7*(-3)-8*(3)], we get the following:
17/4 = c
[Equation 7*(2)-8] is as follows:
-2a+c = 9/4
Substituting the value of c in the above equation yields the following:
2a = 8/4
Further simplifying yields the following value for a:
a = 1
[Equation 7] with a and c substituted is as follows:
33/4 = 1 + b + 17/4
Solving for b yields the following:
b = 12/4
Further simplifying yields the following value for b:
b = 3
Thus, r(n) = 4/(4n^2 + 12n + 17)
As a sanity check, we verify that this formula works for r(4).
From the recurrence relation, r(4) = 4/129. The closed-form formula for r(n) yields 4/(64 + 48 + 17), which is nothing but 4/129. I’m confident that the closed-form formula will work for any positive integral value of n in the context of this puzzle.
Conclusion: The puzzle stands solved, r(n) = 4/(4n^2 + 12n + 17), and I can finally going to bed.
This puzzle was obtained from brilliant.org’s Puzzle of the Week for the week of February 12, 2018. I am not related to brilliant.org in any way, have not been compensated by them in any form, and they didn’t tell me to write this article. This particular puzzle is the fourth one in the advanced section.
