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# NTHCIR - Editorial

Tester: Mugurel Ionut Andreica
Editorialist: Lalit Kundu

Medium-Hard

# PROBLEM:

We are given radius of circles $C_1$, $C_2$, $C_3$ and $C_4$ in the following image:

Now, we are going to draw circles $C_5$, $C_6$ and so on in such a way that $C_n$ touches circles $C_1$, $C_2$ and $C_{n-1}$. Like as shown in the following image:

Given upto $10^7$ values of $n$, find sum of radius of circle $C_n$ for all $n$.

# QUICK EXPLANATION:

========================
Using Descartes' Theorem, if we have $3$ existing circles of curvature(inverse of radius) $a_1$, $a_2$, $a_3$, we can draw a $4^{th}$ circle of curvature $a_4$, touching all three circles such that $a_4 = a_1 + a_2 + a_3 \pm 2 \sqrt{a_1*a_2 + a_2*a_3 + a_1*a_3}$, where $a_4$ is negative if it represents a circle that circumscribes the first three.

Now, $n^{th}$ circle in our problem is formed by finding the $4^{th}$ circle touching circles $C_1$, $C_2$ and $C_{n-1}$. So, we can define our recurrence, solving which we get general term for the radius of $n^{th}$ circle.

# EXPLANATION:

================

One of the ways to solve this problem is via Descartes' Theorem. First, lets define what curvature of a circle is.

The curvature of a circle with radius $r$ is defined as $\pm \frac{1}{r}$. The plus sign here applies to a circle that is externally tangent to the other circles, like the red circle is externally tangent to the three black circles in the image. For an internally tangent circle like the big red circle, that circumscribes the other circles, the minus sign applies.

Descartes' Theorem states that if four circles are tangent to each other at six distinct points, and the circles have curvatures $a_i$ (for $i = 1, ..., 4$), then:

$(a_1+a_2+a_3+a_4)^2=2(a_1^2+a_2^2+a_3^2+a_4^2)$.

When trying to find the radius of a fourth circle tangent to three given kissing circles(circles that touch each other mutually), the equation is best rewritten as:

$a_4 = a_1 + a_2 + a_3 \pm 2 \sqrt{a_1*a_2 + a_2*a_3 + a_1*a_3}$

The $\pm$ sign reflects the fact that there are in general two solutions. One solution is positive and the other is either positive or negative; if negative, it represents a circle that circumscribes the first three (as shown in the diagram above).

Now, let's see how we employ this theorem to find solution to the problem.

We know that $C_n$(for $n \ge 4$) is formed as a $4^{th}$ circle tangent to three kissing circles $C_1$, $C_2$ and $C_{n-1}$, where $C_1$ is being touched by all circles internally. And also, the circle $C_n$ will touch all other circles externally, so we are interested in positive curvature of $C_n$. And, curvature of $C_1$ is $-\frac{1}{r_1}$.

So, we can write using Descartes' Theorem that $a_n = a_1 + a_2 + a_{n-1} + 2 \sqrt{a_1*a_2 + (a_2 + a_1)*a_{n-1}}$. Now, we can employ this recurrence to find $n^{th}$ circle, in $O(n)$. This will be enough to pass subtask 1.

For subtask 2, we need a constant time solution for each $n$, that means we need to arrive at a closed form of the $n^{th}$ term of recurrence. Let's see how we can solve this recurrence. Basically, our recurrence is:

$a_n = c_1 + a_{n-1} + 2 \sqrt{c_2 + c_1 a_{n-1}}$,
where $c_1 = a_1 + a_2$ and
$c_2 = a_1 * a_2$

Here, we substitute $x_n$(a new recurrence) as $\sqrt{c_2 + c_1 a_{n}}$, which implies that
$a_n = \frac{x_n^2 - c_2}{c_1}$ --------Eqn (1)

So, our recurrence is now $a_n = c_1 + a_{n-1} + 2 x_{n-1}$. In this equation, substitute $a_{n-1}$ and $a_n$ using equation $1$ to get,

$x_n^2 = x_{n-1}^2 + 2c_1x_{n-1} + c_1^2$, which is basically $x_n^2 = (x_{n-1}+ c_1)^2$. Now, this has greatly been simplified to $x_n = x_{n-1} + c_1$, whose general term can be written as $x_n = nc_1 + C$, where $C$ is an arbitrary constant, which depends on $x_1$.

Now, to find $a_n$ we substitute general term for $x_n$ in equation $1$, which yields $a_n = \frac{(nc_1 + C)^2 - c_2}{c_1}$.

Now, we know the value of $a_3$ as $\frac{1}{r_3}$, so we can find value of $C$ by substituting $n$ as $3$ in the general form of $a_n$. As you might note, it will give us two values for $C$ because its a quadratic in $C$, so, we take the value which gives us correct value for $a_4$ whose value we already know as $\frac{1}{r_4}$.

Now, our solution for a single $n$ is constant time.

# ALTERNATE SOLUTION:

========================

There is another solution using the concept of inversion of a plane. Let's see this concept in detail. We are going to transform all points in an $x$-$y$ plane(say plane 1) to another $x$-$y$ plane(say plane 2). For each point $(x, y)$ in plane 1, we map it to $\frac{x}{r^2}, \frac{y}{r^2}$, where $r$ is distance of point $(x, y)$ from origin i.e. $\sqrt{x^2 + y^2}$. If we consider point $(x, y)$ in polar coordinates as $r e^{i \theta}$, then we map it to $\frac{1}{r} e^{i \theta}$ in the plane 2. Transformation from plane 2 to plane 1 is simple, we just have to apply the same transformation again for all points in plane 2.

Now, let's see how it affects simple figures: circle and lines. Here I am going to summarise the results, you can easily derive these results by considering equations of lines and circles(doing in polar coordinates will be easier).

• All the circles which pass through origin in plane 1 will be mapped to straight lines in plane 2. A circle with center at $(r, 0)$ and radius $r$ will be mapped to line $x = \frac{1}{2 r}$ in plane 2.

• All the straight lines which do not pass through origin in the first plane will be mapped to circles which pass through origin in the second.

• All the circles which do not pass through origin in the first plane will be mapped to some other circles in the second plane.

• All the straight lines which pass through origin in the first plane will be mapped to the same lines in the second. Also, note that if there are two points on such a line where one point is at a distance $r_1$ from origin and another at distance $r_2$, where $r_2 \gt r_1$, then distance between them in plane 1 is $r_2 - r_1$ and distance between these points in transformed plane will be $\frac{1}{r_1} - \frac{1}{r_2}$.

Now, let's see how we employe this transformation to solve our problem. First, we'll choose a suitable origin in plane 1(plane where all our circles are present). Now, we see that all circles are tangent to circles $C_1$ and $C_2$ and we remember that we can transform circles passing through origin to straight lines, so, we choose the point where $C_1$ and $C_2$ touch as the origin. Now, we also assume that their centers lie on $y$-axis at points $R_1$ and $R_2$ respectively.

Now if we transform the plane, then circles $C_1$ and $C_2$ will become straight lines which are parallel to each other. All the circles $C_3$, $C_4$, $C_5$ $\cdots$ will become some other circles in the transformed plane. But the fact that they all are touching $C_1$ and $C_2$ before transformation means that they all touch the two straight lines which are parallel to each other after transformation, which makes all their radii equal in the transformed plane(this radius is equal to $\frac{1}{4R_1} - \frac{1}{4R_2}$). See the following figure.

So, the equations of lines $C_1^{\prime}$ and $C_2^{\prime}$ are $y = \frac{1}{2 R_1}$ and $y = \frac{1}{2 R_2}$ respectively.

Now, to find the radius of any $n^{th}$ circle in plane 1, we first find coordinates(in plane 2) of two points on the circle and we transform these points back to plane 2 and get the distance between them as the diameter.

We already know the $y$-coordinate of the centre of circles $C_3^{\prime}$, $C_4^{\prime}$ and so on. We need to find their $x$ coordinate now. If we have $x$-coordinate of the center of circle $C_3^{\prime}$, then we can easily find centre of any circle and we already know the radius of circles in plane 2, so we can also find the coordinates of a point on the circumference, which we'll transform back.

Let's see how we find $x$-coordinate of center of $C_3^{\prime}$ in plane 2. Keep referring to image below while reading this for more clarity. We know that a line passing through origin in plane 1 remains same in plane 2, so we draw a line $Z_1$ in plane 1 passing from origin and center of circle $C_3$. Now, this line in plane represented as $Z_1^{\prime}$ will also pass through center of circle $C_3^{\prime}$ and origin of plane 2. If we know distance of center of $C_3^{\prime}$ from origin somehow, we can calculate its $x$-coordinate also, since we already have $y$-coordinate.

We now, try to relate distances in plane 2 with distances in plane 1. Say, the distance $d$ = distance between origin and center of $C_3^{\prime}$. So, circle $C_3^{\prime}$ cuts line $Z_1^{\prime}$ at two points, distance between these points in plane 1 must be the diameter of $C_3$ i.e. $2R_3$ because the same line intersects circle $C_3$ at same two points. Distance of these two points from origin in plane 2 can be written as $d - r_3^{\prime}$ and $d + r_3^{\prime}$. Now, if we transform them to plane 1 and see distance between them, we can say that:

$2 r_3 = \frac{1}{d - r_3^{\prime}} - \frac{1}{d + r_3^{\prime}}$

In above equation, we can find the $d$ as all other variables are known. And using value of $d$ we can find $x$-coordinate of center of $x_3^{\prime}$ $=$ $C_3^{\prime}$ as $\pm \sqrt{d^2 - y^2}$. So, center of $n^{th}$ circle will be $x_3^{\prime} + 2(n-3)R_3$ or $x_3^{\prime} - 2(n-3)R_3$, we can now check which one to use by finding radius of $C_4$ and match with given value in input.

Now, our problem is solved. You can do the calculations yourself to find the closed formula.

# COMPLEXITY:

Constant per query.

# AUTHOR'S, TESTER'S SOLUTIONS:

This question is marked "community wiki".

3.0k93163187
accept rate: 12%

19.7k350498541

 4 There is one more recurrence relation which can be formed... as we know ..C(N)=C(1)+C(2)+C(N-1) +/- 2*SQRT(REMAINING PART :D) we can also see the complement circle for n th circle will be (n-2)th circle... so if C(N)=C(1)+C(2)+C(N-1) + 2*SQRT(REMAINING PART :D) THEN C(N-2)=C(1)+C(2)+C(N-1) - 2*SQRT(REMAINING PART :D) ADDING BOTH EQUATIONS WE GET : C(N)+C(N-2)=2C(1)+2C(2)+2C(N-1).. answered 13 Jul '15, 15:43 2★coolsduy 165●9 accept rate: 25% What do you mean by complement circle? Also how to do you prove that complement of nth circle is (n-2)th circle (13 Jul '15, 16:12) likecs6★ if u see by descartes theorem...for example initially u are given 4 circles... 1 outer C1 and three inner circles C2 C3 C4....now to make a new circle C5...u will put curvatures of C1,C2,and C4 in descartes equation which will provide...two solutions..as it is quadratic in nature...therefore one solution is the new circle..what will be the other solution...the one which was already touching C2 C1 and C4...and what was that? C3 ryt...therefore ..C(n) and C(n-2) are complement circles...u can see this by drawing :D ...though the recurrence relation u guys made was easier to solve than this one. (13 Jul '15, 16:37) coolsduy2★ 1 Yes, this is the recurrence I used in the tester's solution. Then I simply found the general term of the linear recurrence. (13 Jul '15, 16:57) thanks, well explained. (13 Jul '15, 22:58) likecs6★
 2 Even in my JEE I never came across even such terms "Apollonian Gaskets","Soddy Circles","Descartes theorem","Inversions"!! :/ answered 13 Jul '15, 16:02 142●6●18 accept rate: 11%
 2 it is similar to this problem whose solution is here. But what makes it interesting is the fact that the inversion about the circle at origin can be shifted by some amount delta .So, you have to use the radii of the circles r3 and r4 for calculating the shift and backtranslate from the smaller circles to find the radii of the nth circle. The formulas can be found here.. My submission is here.. It is a really cool concept. answered 13 Jul '15, 21:43 19●2 accept rate: 0% The formulas(8) and (9) will be sufficient as mentioned in the wolfram link. (13 Jul '15, 21:48)
 0 Why dont we square on both sides or in other words use (k_n + k_n-1 + k_1 +k_2)^2 = 2((k_n)^2 + (k_n-1)^2 + (k_1)^2 + (k_2)^2)...and then solve recursively. By the way this question was like the "Mother Of Simplification". I couldnt believe my own eyes when it boiled down to such a small formula which is kn=k4 + (n-4)((k4-k3) + (n-3)*(k1+k2)) where n>4. answered 13 Jul '15, 15:52 80●5 accept rate: 10% Could you please provide a link to image or any thing related to how you found out the solution to the recurrence. I am stuck at some steps...or any strategy that i should try. (16 Jul '15, 10:15) apptica5★
 0 Instead of solving the recurrence, what I noticed was that k[i] (curvature of i-th circle) had a sequence such that the difference was in arithmetic progression. In this way too, we can calculate the curvature of n-th circle in O(1). answered 13 Jul '15, 15:54 6★ketanhwr 1.9k●3●18●44 accept rate: 15%
 0 i got wa in the 2nd task of 2nd subtask ....probably due to precision ....i used the formula kn=(n-3)^2 * (k1+k2) + k3 + 2(n-3)sqrt(k1k2+k2k3+k1*k3) in order ...what precision error could possibly pop out ? answered 13 Jul '15, 15:57 3★anin_217 1 accept rate: 0%
 0 What a beautiful question..... got the same recurrence but was not able to prove it.... Also, Codechef please give the 2nd test case (1st subtask) values and solution as many got WA in this (including me).... Later found matrix while surfing net, which can also help to solve for nth radius using Matrix exponentiation. answered 13 Jul '15, 16:08 6★likecs 3.7k●22●79 accept rate: 9% Second test case of first subtask was probably wrong due to floating point errors. I also got WA on this test case for a long time until I finally fixed my floating point calculations and made them as precise as possible. (13 Jul '15, 17:27) Why can we write xn=xn−1+c1 as xn=nc1+C ? (14 Jul '15, 21:12)
 0 I was wondering why this code didn't work for the first subtask? It uses Descartes formula, yet it still gave WA. Can anyone tell me where is the mistake? LL n, p, m, b; double r1, r2, r3, r4, t1, t2, res; vector R; int main() { //freopen("test.in","r",stdin); //freopen("test.out","w",stdout); int T; scanf("%d", &T); scanf("%lld%lld%lld%lld", &n, &p, &m, &b); scanf("%lf%lf%lf%lf", &r1, &r2, &r3, &r4); R.PB(r1); R.PB(r2); R.PB(r3); R.PB(r4); r1 = 1.0 / r1; r2 = 1.0 / r2; rb(i, 4, 2000) { r3 = r4; r3 = 1.0 / r3; t1 = r2 + r3 - r1; t2 = (r2 * r3) - r1 * (r2 + r3); r4 = t1 + 2.0 * sqrt(t2); r4 = 1.0 / r4; R.PB(r4); } res = 0.0; while(T--) { //scanf("%d%d", &n, &q); n = (n * p) % m + b; res += R[n - 1]; //printf("%lld\n", res + K[n][k]); } printf("%.9lf\n", res); return 0;  } answered 13 Jul '15, 17:15 4★bondoc73 106●2●4●10 accept rate: 10% What does rb(i, 4, 2000) do? If it loops i from 4 to 2000, that was probably the problem because for Subtask 1, I think N_i could be as big as 3000. (13 Jul '15, 17:30) rb(i, 4, 2000) compute the radius of all circles from 5 till 2000. I also tried rb(i, 4, 2000000) and I still got wrong answer for subtask 1. I still couldn't find the problem. (14 Jul '15, 00:25) bondoc734★ you have to print exactly upto 6 decimal places... (14 Jul '15, 09:46)
 0 please tell me why this did not pass even first sub task? #include #include #include typedef long long int ll; typedef long double ldf; using namespace std; ll mulmod(ll a, ll b, ll c) { ll x = 0; ll y = a%c; while(b>0) { if(b%2 == 1) x = (x+y)%c; y = (y*2)%c; b/=2; } return x%c; } int main() { ll t; scanf("%lld",&t); ll n,p,m,b; scanf("%lld%lld%lld%lld",&n,&p,&m,&b); ll r1,r2,r3,r4; scanf("%lld%lld%lld%lld",&r1,&r2,&r3,&r4); ldf k1,k2,k3,k4; k1 = 1.0/r1; k2 = 1.0/r2; k3 = 1.0/r3; k4 = 1.0/r4; ldf ans = 0; for(ll x=1;x<=t;x++) { n = mulmod(p,n,m) + b; if(n==1) { ans += 1.0/k1; continue; } if(n==2) { ans += 1.0/k2; continue; } if(n==3) { ans += 1.0/k3; continue; } if(n==4) { ans += 1.0/k4; continue; } ldf z = k4; for(ll i=5;i<=n;i++) z = k2 - k1 + z + 2.0*sqrt(k2*z - k1*(k2 + z)); ans += z; } ll ans1 = ans*10000000; ans = ans1/10000000.0; printf("%.6Lf",ans); return 0; }  answered 13 Jul '15, 19:05 226●2●9 accept rate: 9% Precision takes a toll when you are taking square roots so many times! (13 Jul '15, 19:23) yash_155★ @yash_15 i don't get you, i tried casting the sqrt function to long double, didn't work (13 Jul '15, 19:44)
 0 Were test cases weak? I used pappus chain n got 75 . Pappus chain is a special case I guess answered 13 Jul '15, 19:14 16●1 accept rate: 14%
 0 Did someone manage to get accepted this problem using solution no.2 form editorial? I'm failing so far. Maybe someone could help me? Please, take a look at my submission, e.g. function solve2_editorial. Maybe, someone got it accepted using solution no.2? Could you please share your code? p.s. Admins, Setter code is not available! Please, fix it! answered 13 Jul '15, 19:41 4★k0stia 41●6 accept rate: 0%
 0 Please help me with my code. My code fails on second test file in sub task 1. :/ Thanks in Advance. View my Code here! answered 13 Jul '15, 19:48 4★lohit_97 342●8 accept rate: 4%
 0 How to proceed after an=((n*c1+C)^2-c2)/c1 to obtain C??? for eg for the test case 1 1 2 1 5 6 1 1 2 we get two values of C=-1.684,-3.316 by substituting n=3 and a3=1/r3 but none of these values hold for n=4 & a4. plzz explain answered 13 Jul '15, 21:12 1 accept rate: 0%
 0 @vijay_cipher could you provide more details on your solution? I am struggling with this single line: yl=(.25*((1.0/r4)-(1.0/r3)))-rs; As far as I understand this is the center of left most circle (C'_3) after inversion. But do not understand how you derived this formula. Could you provide more details on it? Thanks. answered 14 Jul '15, 00:11 4★k0stia 41●6 accept rate: 0%
 0 If you have the curvatures of 4 circles you can get the curvature of the 5th circle easely by S*V, where Ci=1/Ri, S is a matrix S=(1 0 0 0; 0 1 0 0; 0 0 0 1; 2 2 -1 2) and V=(C1 C2 C3 C4) initially. So we need the 4th line of S^n that is n(n+1) n(n+1) -n n+1. n=1 => C5 If you have 4 circle curvatures then u can multiply the vector of the curvatures with 4 matrices and generate all the possible circles in an Apollonian gasket. S is actually S3 and corresponds to a new circle tangent to the first, third and fourth circles.  answered 15 Jul '15, 12:10 4★jackal02 66●2 accept rate: 25%
 0 I used a simpler recursion to find the radius: note that circles $C_n$ and $C_{n+2}$ are both tangent to the triplets $C_1, C_2, C_{n+1}$, so $\dfrac{1}{R_n}$ and $\dfrac{1}{R_{n+2}}$ both satisfy the Descartes theorem equation for $C_1, C_2, C_{n+1}$. One easily gets that the sum of the roots of the equation is $2(-1/R_1 + 1/R_2 +1/R_{n+1})$. We know one of the roots is $1/R_n$ and the other is $1/R_{n+2}$, so $\dfrac{1}{R_{n+2}}= 2 \left( -\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_{n+1}} \right ) - \dfrac{1}{R_n}$. From here it's easy to see that $\dfrac{1}{R_n}$ is a quadratic in $n$ and we plug in the given values to find the coefficients. answered 16 Jul '15, 10:22 157●1●8 accept rate: 10%
 0 @bondoc, i also had the same concept and problem earlier. But according to the comments mentioned in the problem during the contest, the radius of the circle can increase after some interval. For similar figure you can refer to figure of Apollonian gasket. Also, see the first solution of lebron (link-http://www.codechef.com/viewsolution/7391329) he also used the same concept but with one difference as soon as the radius becomes so small that it is negligible, he starts to takes the reverse series again. May be this helps. answered 16 Jul '15, 11:25 6★likecs 3.7k●22●79 accept rate: 9%
 0 Nice! Thanks :) answered 16 Jul '15, 19:44 1 accept rate: 0%
 0 Can anyone post a nice link to better understand the concept of inversion ? There aren't many tutorials about it clubbed with the programming category. Is it important? answered 16 Jul '15, 19:44 291●2●3●19 accept rate: 10%
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question asked: 07 Jul '15, 02:39

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last updated: 29 Jul '15, 17:07