AoPS Academy
events 
that are each continuous and last on a certain time interval. Each event has a weight 
However, one can only choose to attend activities that do not overlap with each other. The goal is to maximize the sum of weights of all activities attended. Prove or disprove that the following algorithm allows for an optimal selection:
consider 
the sum of 
over all 
such that 
and are not compatible.
If there are multiple such events, delete the event with the minimal weight.
are 
such that 
and 
.
be a triangle with 
. Let 
be the reflection of 
across the line 
and be the reflection of 
across the line 
. Let 
and 
intersect at 
. Prove that the orthocenter of triangle coincides with the circumcenter of triangle 
.
be real numbers such that 
and 
. Find the largest possible value of 
.
characters from the text, the characters will be the same.
.
is the number of characters in the alphabet, 
is the number of times the 
th of the alphabet appears in the plaintext, and 
is the number of letters in the plaintext. letters from a group of letters? It is of course, 
which is equal to 
. This is the denominator. To calculate the numerator, we first calculate the number of ways to pick a's from our plaintext and add that to the number of ways to pick b's from our plaintext, and so on. If the number of a's in our plaintext was , then the number of ways to pick a's is 
, this is equal to 
as we have shown before. This numerator and denominator gives us 
, the 
's cancel giving us our desired formula:
.
should ALWAYS be encrypted as the letter . In a Vigenere cipher this does not occur. Depending of 's position in the plaintext, could be encrypted as one of several letters. If the keyword has length 
, then could be encrypted as 
. This is not a one to one correspondence so frequency analysis does not work. in normal English was . If the key word was of length 
, the frequency of in the cipher text would be as well. But if the key word was of length , then the frequency of 
and the frequency of 
. Basically frequency analysis works if there is one alphabet that corresponds to another alphabet in a 1 to 1 correspondence.
.
and the letters of our key transformed into numbers are 
.
we use 
where 
. This accounts for the fact that the key is repeated over each letter of the plaintext. We use mod because is the length of the keyword.
the corresponding ciphertext number to .
where 
. changes changes as well. This is what makes the Vigenere cipher a much better code.
th number of our plaintext, , to find the equivalent key letter we take 
. So we take the th keyword number which is 
.th number of our plaintext, 
to find the equivalent key letter we take 
. So we take the nd keyword number which is 
.
. So our 0th ciphertext number is 23.
. So our 5th ciphertext number is 1.
choices because we can choose any letter of the alphabet. Let us say we chose "R".
choices. Let's say we chose "E". Now for the next question mark we can't chose the letter "R" or "E" so we have 
choices. By now you should see the pattern, we have 
choices. That is equivalent to factorial.
.
is the key. is the letter being encrypted and is the encrypted letter. The variables p and c are used to represent the letter being encrypted because in cryptography we refer to the original message as the 'plaintext' and the encrypted message as the 'ciphertext'., represents a number because it is the key. But and are actually letters. We need to convert them into numbers. This is very simple. Represent A by 0, B by 1, C by 2 ... Z by 25. letters. In terms of numbers we are just adding to to get .
.
and 
then 
which is a number we cannot convert to a letter. This problem occurs because of the 'wrapping around' from Z to A. To fix this we can use modular arithmetic. If you don't know what this is try googling it. We will almost always be working in mod 26 because there are 26 letters in the alphabet. Our new equation would be:
we should subtract it.
touches the sides 
in 
, respectively, and 
are the midpoints of 
, respectively. Prove that the centers of the inscribed circle and of the circles around 
and are collinear.
be the excenter of 
respectively. 
be the centroid of 
.
be the circumcenter of 
is homothetic.Let 
be the center of a homothety takes 
to 
.We'll get 
.So We'll conclude 
Our proof is completed
mean 
are collinear
and 
be the incenter and excentral triangle of . Also, Let 
be the Bevan point of . Then lies on line 
, where 
is the circumcenter of .
. Thus, we get that and are homothetic, and their homothety center lies on the line joining their centers, i.e. line 
. This gives that line is the Euler line of , proving that the nine point center of (i.e. the circumcenter of 
) lies on line .
touches the sides in , respectively, and are the midpoints of , respectively. Prove that the centers of the inscribed circle and of the circles around and are collinear.. Then the circumcircle of the triangle will be the nine points circle of 
(the nine points circle of is the circumcircle of the triangle )., and the circumcenter of are collinear. is the euler line of the contact triangle of the triangle .
be circumcenter of 
respectively.
so 