Best bid given the other's bid
by fancyfairy, Mar 30, 2025, 10:04 PM
Consider sets 
(player) and 
(items), with valuations given by matrix
![$
V = \left[\begin{matrix}
v_{1a} & v_{1b} & v_{1c} & v_{1d} \\
v_{2a} & v_{2b} & v_{2c} & v_{2d}
\end{matrix}\right]
$](//latex.artofproblemsolving.com/d/4/6/d465348852f12739e1fc2a04a9c2e5fdad0dbcb4.png)
where![$v_{ij} \in [0, \infty]$](//latex.artofproblemsolving.com/7/8/8/788bffd362b46a318a2cb0703a9e2ae77089e4e7.png)
.
A player’s total valuation is the sum of item values won, with half-value for half-won items.
Given player
's bid, player 
's bid 
is strictly better than another possible bid 
of his if the total valuation derived from is strictly more than that derived from 
.
Each player simultaneously bids on items with a bid vector![$B_i = [b^i_1, b^i_2, b^i_3, b^i_4]$](//latex.artofproblemsolving.com/7/b/2/7b233ca3e9056aa374e937b1133cd9c81726f407.png)
such that ![$b^i_j \in [0,1]$](//latex.artofproblemsolving.com/b/9/6/b96c2a7618767934c8a44d9238c485fed39b8510.png)
and 
. The highest bidder wins an item; ties split ownership equally.
Given valuation matrix
, can we always find bid vectors 
and 
such that given , the second player can not submit a bid strictly better than and vice-versa?
(player) and 
(items), with valuations given by matrix![$
V = \left[\begin{matrix}
v_{1a} & v_{1b} & v_{1c} & v_{1d} \\
v_{2a} & v_{2b} & v_{2c} & v_{2d}
\end{matrix}\right]
$](http://latex.artofproblemsolving.com/d/4/6/d465348852f12739e1fc2a04a9c2e5fdad0dbcb4.png)
where
![$v_{ij} \in [0, \infty]$](http://latex.artofproblemsolving.com/7/8/8/788bffd362b46a318a2cb0703a9e2ae77089e4e7.png)
.A player’s total valuation is the sum of item values won, with half-value for half-won items.
Given player
's bid, player 
's bid 
is strictly better than another possible bid 
of his if the total valuation derived from is strictly more than that derived from 
.Each player simultaneously bids on items with a bid vector
![$B_i = [b^i_1, b^i_2, b^i_3, b^i_4]$](http://latex.artofproblemsolving.com/7/b/2/7b233ca3e9056aa374e937b1133cd9c81726f407.png)
such that ![$b^i_j \in [0,1]$](http://latex.artofproblemsolving.com/b/9/6/b96c2a7618767934c8a44d9238c485fed39b8510.png)
and 
. The highest bidder wins an item; ties split ownership equally.Given valuation matrix
, can we always find bid vectors 
and 
such that given , the second player can not submit a bid strictly better than and vice-versa?polynomial
by hanzo.ei, Mar 30, 2025, 3:12 PM
Given a polynomial 
satisfying 
, 
, and for 
(
) positive real numbers 
. Prove that there exists a strictly increasing sequence of real numbers 
such that
![\[
\sum_{i=1}^{n} \frac{k_i}{P'(a_i)} = \sum_{i=1}^{n} k_i.
\]](//latex.artofproblemsolving.com/4/4/e/44e5eb17b17531e36c362e062471d9bdffc8144c.png)
satisfying 
, 
, and for 
(
) positive real numbers 
. Prove that there exists a strictly increasing sequence of real numbers 
such that![\[
\sum_{i=1}^{n} \frac{k_i}{P'(a_i)} = \sum_{i=1}^{n} k_i.
\]](http://latex.artofproblemsolving.com/4/4/e/44e5eb17b17531e36c362e062471d9bdffc8144c.png)
My problem
by hacbachvothuong, Mar 29, 2025, 10:10 AM
Let 
be positive real numbers such that 
. Prove that:
be positive real numbers such that 
. Prove that:
![\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]](http://latex.artofproblemsolving.com/c/f/8/cf855cb087102024a50d41f49c67e7fd1ef22767.png)
Bishops and permutations
by Assassino9931, Feb 29, 2024, 7:57 PM
Let 
be a positive integer. Initially, a bishop is placed in each square of the top row of a 
chessboard; those bishops are numbered from
to 
from left to right. A jump is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row.
Find the total number of permutations
of the numbers 
with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order 
, from left to right.
Israel
be a positive integer. Initially, a bishop is placed in each square of the top row of a 
chessboard; those bishops are numbered from
to 
from left to right. A jump is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row.Find the total number of permutations
of the numbers 
with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order 
, from left to right.Israel
This post has been edited 1 time. Last edited by Assassino9931, Mar 4, 2024, 10:59 AM
Prove that AY is tangent to (AEF)
by geometry6, Aug 11, 2021, 9:03 AM
Let 
be an arbitrary interior point of 
, and 
, 
, 
intersect 
, 
, 
at 
, 
, 
, respectively. Suppose that 
be the midpoint of , 
and 
intersect at 
, 
intersects at 
, and 
intersects at 
. Show that 
is tangent to .
be an arbitrary interior point of 
, and 
, 
, 
intersect 
, 
, 
at 
, 
, 
, respectively. Suppose that 
be the midpoint of , 
and 
intersect at 
, 
intersects at 
, and 
intersects at 
. Show that 
is tangent to .IMO 2016 Problem 1
by quangminhltv99, Jul 11, 2016, 6:23 AM
Triangle 
has a right angle at 
. Let 
be the point on line 
such that 
and lies between and 
. Point is chosen so that 
and 
is the bisector of 
. Point is chosen so that 
and 
is the bisector of 
. Let be the midpoint of . Let be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.
has a right angle at 
. Let 
be the point on line 
such that 
and lies between and 
. Point is chosen so that 
and 
is the bisector of 
. Point is chosen so that 
and 
is the bisector of 
. Let be the midpoint of . Let be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.This post has been edited 2 times. Last edited by sseraj, Jul 11, 2016, 4:41 PM
Prove that $\angle FAC = \angle EDB$
by micliva, Apr 18, 2013, 7:06 PM
Points and are given on side of convex quadrilateral 
(with closer than to ). It is known that 
and 
. Prove that 
.
M. Smurov
and are given on side of convex quadrilateral 
(with closer than to ). It is known that 
and 
. Prove that 
.M. Smurov
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