Hi I'm working on my scholarship exam practice but I got stuck at my last question, could you please have a look? I'll write every part of question so you know the big picture. I have marked the problematic question in bold below. This question only assumes knowledge in high school or pre-university level.  Draw a circle $C$ on a plane. Put $n$ distinct points on the circumference of $C$. Join any two of the $n$ points with a chord. Suppose that no three chords intersect at any one point inside $C$. Let $c_n$ be the number of such chords, $i_n$ that of intersections of the chords inside $C$, and $r_n$ that of regions in $C$ bounded by an arc and/or some chords. Fill in the blanks with the answers to the following questions. (1) Find $c_1, c_2, c_3, c_4, c_5, c_6$ and write their values in this order.  The answer of this part checked with answer key is $0, 1, 3, 6, 10,$ and $15$ by drawing 6 different circles with n dots from 1 to 6 each circle, connecting each dot and counting number of chords.  (2) Find $i_1, i_2, i_3, i_4, i_5, i_6$ and write their values in this order.  The answer of this part checked with answer key is $0, 0, 0, 1, 5,$ and $15$ by counting intersections of chords.  (3) Find $r_1, r_2, r_3, r_4, r_5, r_6$ and write their values in this order.  The answer of this part checked with answer key is $1, 2, 4, 8, 16,$ and $31$ by counting the regions.  (4) Express $r_n$ with binomial coefficients in terms of $n$. (For binomial coefficients ${m}\choose{k}$, note that ${m}\choose{k}$$= 0$ if $m&lt;k$.)  I got stuck at this question since I got nowhere near the answer key. My approach is below: ${n}\choose{k}$$=\frac{n!}{(n-k)!\cdot k!}$ Pick $n=2$ and $n=3$ and substitute the values from (3) into the equation, ${2}\choose{k}$$=\frac{2!}{(2-k)!\cdot k!}=2$ ${3}\choose{k}$$=\frac{3!}{(3-k)!\cdot k!}=4$ $\frac{2!}{(2-k)!\cdot k!\cdot2}=\frac{3!}{(3-k)!\cdot k!\cdot4}$ And $k$ will be weirdly equal to $\frac{3}{2}$ leading to binomial coefficient ${n}\choose{3/2}$, which is incorrect. The answer key provided is $1 +$${n}\choose{2}$$+$${n}\choose{4}$. How can we get to this answer? And, how is my method faulty?
124285
Assuming there are n points on a circle, then there are $C_n^2$ chords. If any three chords don't intersect in circle. How to show the circle will be divided into $1+ C_n^2 + C_n^4$ domains?  Picture below is a example of 5 points on circle.
82678
By placing 27 squares with side $1$ in a circle with radius $2$, such that each square is entirely inside the circle, prove that there exists a point that 3 squares share (inside of their area).    I thought about dividing the circle into 9 equal areas and using the pigeonhole principle, but got stuck there. Any hint will help.
114324
Prove that having 6 points in the interior of a square of side length 3, we can choose 2 of them so that the distance between them is less than 2.  Looks obvious, but I can't get a rigorous demonstration.   I tried to cover the square with 6 circles of radius 1, having the centers inside the square, to prove they must intersect at least once.  Note:  This partition leads to solution.  The rectangles on the first row: $3/2 * (3 - \sqrt3) $   The rectangles on the second row: $1 * \sqrt3 $   The 5 zones are so that the distance between any 2 points inside the same zone is less than 2.
64365
Problem statement:  Given $z_1,z_2,z_3,z_4$ different points of $\overline {\mathbb C}$, we define the cross ratio $(z_1,z_2,z_3,z_4)$ as $(z_1,z_2,z_3,z_4)=\dfrac{z_1-z_2}{z_1-z_4}\dfrac{z_3-z_4}{z_3-z_2}$.  Note that $(z_1,z_2,z_3,z_4)$ is the image of $z_1$ under the Möbius transformation $T$ such that $T(z_2)=0$, $T(z_3)=1$, $T(z_4)=\infty$.  a) Prove that if $T \in \mathcal H$ then $(T(z_1),T(z_2),T(z_3),T(z_4))=(z_1,z_2,z_3,z_4)$.  b) Show that $z_1,z_2,z_3,z_4$ lie in a line or circle if and only if $(z_1,z_2,z_3,z_4) \in \mathbb R$  My attempt at a solution:  For a), using the "hint" they give, if $T \in \mathcal H$, I did the following:  If I call $H=(z,,z_2,z_3,z_4)$, I can consider $H \circ T^{-1} (z)$. Note that $H \circ T^{-1}(T(z_2))=0$, $H \circ T^{-1} (T(z_3))=1$ and $H \circ T^{-1} (T(z_4))=\infty$. This means that $(z_1,z_2,z_3,z_4)=H \circ T^{-1}(T(z_1))=(T(z_1),T(z_2),T(z_3),T(z_4))$.   I don't know if my answer is correct, I would like to check it, and if anyone has a better or different answer, he/she is very welcome to post it.  For b) I am lost, for the forward implication, I've tried to show that $(z_1,z_2,z_3,z_4)=\overline {(z_1,z_2,z_3,z_4)}$ or that $arg((z_1,z_2,z_3,z_4))$ is a multiple of $\pi$ but I couldn't conclude anything. I would appreciate some help with this point.
87504
How to find the expected area of the intersection of a set of triangles made up $N$ random points that are picked uniformly inside a circle? The triangles must contain the origin of the circle. If none of the trinagles contain the rigin we pick again $N$ random points until at least one triangle contain the origin and calculate the expected intersection area of all the triangles that contain the origin.
60325
Here is a problem I was given: A lattice point in the plane is a point whose coordinates are both integers. Consider a triangle whose vertices are lattice points (0,0),(a,0) and (0,b), where b≤a. Suppose the triangle contains exactly 74 lattice points in its interior, not including those lattice points on the sides of the triangle. Determine the sums of the areas of all such triangles. The only way that I could see to do this problem was to try all possible values of a, checking to see which values of b gave 74 lattice points. I did this and gave up at a=5 so I suspect there must be an easier way. I was thinking Pick's theorem would be useful but I couldn't think of what to do with it. Can anyone provide a shortcut, or at least just give me the answer? By the way, I know it is less than 1000.
89841
Through a random point inside the circle, we draw $4$ lines with $45$ degrees between each other. Prove, that the total area of odd pieces, equals total area of even pieces.    I suppose there is a solution using integrals, but maybe there is geometric solution as well.
44762
I have a circle centred at $(0,0)$ with radius $r$ and I have $2$ random points $P$ and $Q$ on the circumference of the circle. If $d$ is the distance between $P$ and $Q$, what is the variance of $d$?  My solution: I have taken $P$ with coordinates $(x_1,y_2)$ and $Q$ with coordinates $(x_2,y_2)$  $d=2r \sin(\frac{\theta}{2})$ as it is the equation of chord length.  I have the joint pdf of $x$ and $y$ as :  $f(x,y)=\frac{1}{\pi r^2}$ if $(x,y)$ is on the circle.  $f(x,y)=0$ otherwise.  Will the intergal of $xf(x,y)$ with respect to $x$ and $y$ , give me the expected value of $d$?
70180
Assume that there are $n$ infinitely long straight lines lying on a plane in such a way that no two lines are parallel, and no three lines intersect at a single point. Prove that these lines divide the plane into $\frac{n^2+n+2}{2}$ regions. (Hint: Any two non-parallel straight lines on a plane must intersect at exactly one point.)  Can someone tell me where to start here? I know to prove the base where $n=0$, but don't know how to proceed.  Any help is greatly appreciated!
69890
I have asked this question before  The first guy solved it in a way, but  i would like to know is there any efficient way to do so, because i did the same for N = 13 and N = 15  which took long time.  This problem is from a pen and paper test, ZIO-2018 Can i solve it with an efficient algorithm, well i am pretty beginner in algos, so if there is one kindly descibe it how can i do so?  Thank you
58494
Find the locus of all points inside $\triangle ABC$ such that $PA^{2}+PB^{2}=PC^{2}$. At first, i tried finding a right angled triangle and then tried to go on applying Pythagorean Theorem and finding other trivia but that didn't seem to work. I think the locus is most likely to be some line segment rather than an arc of some circle as that seems to be quite unrelated seen from an Euclidean perspective. Applying Apollonius' Theorem and Stewart's Theorem on some triangles might be the key to this problem although i could not find such triangles. I don't know whether this can be solved by trigonometry or not but since this is a problem from a chapter on euclidean geometry, i am sure that there is some clever way to look at this problem.
87782
Let $u=(6,7), v=(-2, 3), w=(7,0)$. Find the center and the radius of the circumscribed circle of triangle $[u,v,w]$.  My approach would look like this:  (1) Determine the angles of the triangle $[u,v,w]$ via dot products.  (2) If we draw three lines from the the center $o$ of the circle to the three vertices, the triangle $[u,v,w]$ will be made of three equilateral triangles. We can then easily find from (1) all the nine angles of the three triangles.  (3) Determine the heights of the three inscribed triangles, which will be the same for all of the three, and which will be the radius $r$ of the inscribed circle.  (4) Then we can drop a perpendicular line from $o$ to, say, $[v,w]$ in order to find the coordinates of $o$.  But is there a much less tedious way to solve this problem? And do you think my approach is correct?
46654
Given $P(z)=z^6+6z+10$, find how many roots are in each quadrant    I have already seen that $P(z)$ has six different roots, and that none of them are real or of the form $ki$, $k\in \Bbb R$. Since the coefficients of $P(z)$ are real, then it's roots are 3 conjugate pairs, and therefore the possibilities are limited to two.  Defining $f(z)=z^6$, $g(z)=z^6+6z+10$ and applying Rouche in $|z|&lt;2$, $|f(z)-g(z)|&lt;|f(z)|$ I get that all roots are contained in the disc, but how do I go about splitting them into the four quadrants? From what I know about this polynomial, finding zeros in one quadrant is enough to determine the rest, but I'm stuck trying to apply rouche to a quarter circle (basically I'm having trouble finding a suitable f and g)
60872
I've tried solving it, but I'm not familiar with these kind of progressions and I couldn't get to a result. Also, I've found that the result is just $N/2$, but I don't get why would that be the answer, as it doesn't work for $N=1$ or $N=2$. Appreciate your help :)
111664
This question arises from below:  Initially you are given a circle $C_2$ with radius $r=1$. You construct the regualr 3-gon (equilateral triangle) with its incircle as $C_2$, and its circumscribed circle $C_3$. Then use $C_3$ as incircle and repeat this process, so you can construct the regualr $n$-gon with its circumscribed circle $C_n$. Does the limit of the radius of $C_n$ exist as $n\to\infty$ ? If it exists, what is the value?  My attempt  I have worked out that the radius of $C_n$ is  $$r_n=\prod_{i=3}^n\sec\frac{\pi}{i}$$  Since $\lim_{n\to\infty}\sec\frac{\pi}{n}=1$, I concluded that $\lim_{n\to\infty}\frac{r_{n+1}}{r_n}=1$, and the limit of $r_n$ exists. But I cannot find its limiting value. MATHEMATICA says $r_{10^4}\approx8.69574$. Do you have any suggestions on how to proceed?  Thank you in advanced!
102948
If we have two circles in the plane described by $(x_1, y_1, r_1)$ and $(x_2, y_2, r_2)$ we can determine if they are completely disjoint by simply:  $$(x_1 - x_2)^2 + (y_1 - y_2)^2 &lt; (r_1 + r_2)^2$$  Assume this is not the case, we now want to know if one completely overlaps the other.  (That is: if the second circles interior is a subset of the first circles interior.)  If $(x_1, y_1) = (x_2, y_2)$ than we can trivially compare radii, so lets assume their centers are distinct.  The way I have imagined is to create a parametric equation of the line that connects the two centers:  \begin{align} x_p(t) &amp;= x_1 + t(x_2 - x_1) \\ y_p(t) &amp;= y_1 + t(y_2 - y_1) \end{align}  Then we calculate the two line segments (represented as two pairs of t values) where the circles intersect this line.  One is a subset of the other if and only if the corresponding circles are subsets of each other.  Is there a simpler approach I am overlooking?
103815
There are up to $\binom{n}{2} + n + 1$ regions created by a hyperplane arrangement in $\mathbb{R}^2$ containing $n$ hyperplanes.  I want to understand this in a demonstrative way. Each hyperplane splits each region it passes in two regions. A newly added hyperplane therefore creates a new region each time it crosses another hyplerplane or the plane. So the number of regions should be $1$ (plane) $+ \binom{n}{2}$ (number of intersections of $n$ lines). But as I know the answer, I know that I am missing $n$ regions.   Do you know where my mistake is? Why do I have to add the number of lines to get the max. number of regions?