Let and be positive integers and let be a set of points in the plane such that
no three points of are collinear, and
for any point of , there are at least points of equidistant from
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Solution will be updated soon
This year IMO problems !!!!
Problem 1 :
Given triangle the point is the centre of the excircle opposite the vertex This excircle is tangent to the side at , and to the lines and at and , respectively. The lines and meet at , and the lines and meet at Let be the point of intersection of the lines and , and let be the point of intersection of the lines and Prove that is the midpoint of .
Problem 2 :
Let be an integer, and let be positive real numbers such that Prove that
Problem 3 :
The liar’s guessing game is a game played between two players and . The rules of the game depend on two positive integers and which are known to both players.
At the start of the game chooses integers and with Player keeps secret, and truthfully tells to player . Player now tries to obtain information about by asking player questions as follows: each question consists of specifying an arbitrary set of positive integers (possibly one specified in some previous question), and asking whether belongs to . Player may ask as many questions as he wishes. After each question, player must immediately answer it with [i]yes[/i] or [i]no[/i], but is allowed to lie as many times as she wants; the only restriction is that, among any consecutive answers, at least one answer must be truthful.
After has asked as many questions as he wants, he must specify a set of at most positive integers. If belongs to , then wins; otherwise, he loses. Prove that:
1. If then can guarantee a win. 2. For all sufficiently large , there exists an integer such that cannot guarantee a win.
Problem 4 :
Find all functions such that, for all integers that satisfy , the following equality holds:
Problem 5 :
Let be a triangle with , and let be the foot of the altitude from . Let be a point in the interior of the segment . Let be the point on the segment such that . Similarly, let be the point on the segment such that . Let be the point of intersection of and .
Show that .
Problem 6 :
Find all positive integers for which there exist non-negative integers such that
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Thank you and good luck for these delicious problems
These are the set of 5 problems each of 5 marks
The person who get more than 20 points will get a mathematics book on any topic he want…
So try these beautiful problems to test your mathematics abilities and getting new things…
Ways by which you can answer :
I am adding a form in the end of this post , you can answer there ….
Note : You have give solution of your answer too..
The bisectors of the angles and of the meet the sides
and at the points and , respectively.
Assuming that , determine the angle
Given a and be point on side such that ,
Find the value of
If p and q are natural numbers so that
Prove that p is divisible by 1979 .
Find highest degree n of 1991 for which 1991ⁿ divides the number :
Let ƒ(n) denote the sum of the digits of n. Let N = 4444⁴⁴⁴⁴
You can use these symbols to write solutions more conveniently
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Today we will be discussing about 2 important Triangle theorems which have very High applications in field of Geometry
1) Angle Bisector theorem
2) Stewart’s Theorem
Angle bisector theorem :
Because of the ratios and equal angles in the theorem, we think of similar triangles. There are not any similar triangles in the figure as it now stands, however. So, we think to draw in a carefully chosen line or two. Extending AD until it hits the line through C parallel to AB does just the trick:
By AA similarity, . By the properties of similar triangles, we arrive at our desired result:
c/m = b/n
Stewarts Theorem :
Given a triangle with sides of length opposite vertices , , , respectively. If cevian is drawn so that , and , we have that . (This is also often written , a form which invites mnemonic memorization, e.g. “A man and his dad put a bomb in the sink.”)
Applying the Law of Cosines in triangle at angle and in triangle at angle , we get the equations
Because angles and are supplementary, . We can therefore solve both equations for the cosine term. Using the trigonometric identity gives us