MathDB

3

Part of 2017 Saudi Arabia BMO TST

Problems(4)

+ signs between the digits of no 111111 ...111 to be multiple of 30

Source: 2017 Saudi Arabia BMO TST I p3

7/24/2020
How many ways are there to insert plus signs ++ between the digits of number 111111...111111111 ...111 which includes thirty of digits 11 so that the result will be a multiple of 3030?
number theorymultipleDigits
TG=TD, collinear and concurrent wanted, orthocenter, circumcircle related

Source: 2017 Saudi Arabia Mock BMO I p3

7/26/2020
Let ABCABC be an acute triangle and (O)(O) be its circumcircle. Denote by HH its orthocenter and II the midpoint of BCBC. The lines BH,CHBH, CH intersect AC,ABAC,AB at E,FE, F respectively. The circles (IBF(IBF) and (ICE)(ICE) meet again at DD. a) Prove that D,I,AD, I,A are collinear and HD,EF,BCHD, EF, BC are concurrent. b) Let LL be the foot of the angle bisector of ∠BAC\angle BAC on the side BCBC. The circle (ADL)(ADL) intersects (O)(O) again at KK and intersects the line BCBC at SS out of the side BCBC. Suppose that AK,ASAK,AS intersects the circles (AEF)(AEF) again at G,TG, T respectively. Prove that TG=TDTG = TD.
geometrycircumcircleorthocentercollinearconcurrentconcurrency
1,2, 3, 4 around a circle in order.

Source: 2017 Saudi Arabia BMO TST II p3

7/24/2020
We put four numbers 1,2,3,41,2, 3,4 around a circle in order. One starts at the number 11 and every step, he moves to an adjacent number on either side. How many ways he can move such that sum of the numbers he visits in his path (including the starting number) is equal to 2121?
combinatorics
GA+FB=GB+FA and perpendicular wanted, similar triangles , cyclic related

Source: 2017 Saudi Arabia Mock BMO II p3

7/26/2020
Let ABCDABCD be a cyclic quadrilateral and triangles ACD,BCDACD, BCD are acute. Suppose that the lines ABAB and CDCD meet at SS. Denote by EE the intersection of AC,BDAC, BD. The circles (ADE)(ADE) and (BCE)(BC E) meet again at FF. a) Prove that SF⊥EF.SF \perp EF. b) The point GG is taken out side of the quadrilateral ABCDABCD such that triangle GABGAB and FDCFDC are similar. Prove that GA+FB=GB+FAGA+ FB = GB + FA
geometrysimilar trianglescyclic quadrilateralperpendicular