MathDB

2

Part of 2013 Turkey Team Selection Test

Problems(3)

KM.LN=BM.CN

Source: Turkey TST 2013 - Day 2 - P2

4/2/2013
Let the incircle of the triangle ABCABC touch [BC][BC] at DD and II be the incenter of the triangle. Let TT be midpoint of [ID][ID]. Let the perpendicular from II to ADAD meet ABAB and ACAC at KK and LL, respectively. Let the perpendicular from TT to ADAD meet ABAB and ACAC at MM and NN, respectively. Show that ∣KM∣⋅∣LN∣=∣BM∣⋅∣CN∣|KM|\cdot |LN|=|BM|\cdot|CN|.
geometryincenterratiotrigonometrygeometry proposed
19x19 block of 2013x2013 board has at least 21 marked sqrs

Source: Turkey TST 2013 - Day 1 - P2

4/2/2013
We put pebbles on some unit squares of a 2013×20132013 \times 2013 chessboard such that every unit square contains at most one pebble. Determine the minimum number of pebbles on the chessboard, if each 19×1919\times 19 square formed by unit squares contains at least 2121 pebbles.
geometryrectangleinductioncombinatorics proposedcombinatorics
f(x^2) = f(x)^2 - 2xf(x), f(x)=f(x-1), 1<x<y ==> f(x)<f(y)

Source: Turkey TST 2013 - Day 3 - P2

4/2/2013
Determine all functions f:R→R+f:\mathbf{R} \rightarrow \mathbf{R}^+ such that for all real numbers x,yx,y the following conditions hold:
i.f(x2)=f(x)2−2xf(x)ii.f(−x)=f(x−1)iii.1<x<y⟹f(x)<f(y).\begin{array}{rl} i. & f(x^2) = f(x)^2 -2xf(x) \\ ii. & f(-x) = f(x-1)\\ iii. & 1<x<y \Longrightarrow f(x) < f(y). \end{array}
functioninductionalgebra proposedalgebra