This is a number theory problem involving perfect squares and arithmetic sequences. The sequence given is defined by $a_n = 100 + n^2$ for $n = 1, 2, 3, 4$.
★★★★★ Recommended for: Years 12–13, Maths Olympiad students, university applicants preparing for STEP or MAT. bmo 2008 solutions
Find all functions ( f: \mathbbR \to \mathbbR ) such that for all real ( x, y ): [ f(xf(y) + f(x)) = yf(x) + x ] This is a number theory problem involving perfect
Given the complexity, the mark scheme accepted the factorized form and a few explicit pairs: e.g., ( d=2 ) gives ( m=670, n= ? ) Check: ( d=2 ) ⇒ ( m=(2+2008)/3=2010/3=670 ), ( b=2008^2/2=2008*1004=2016032/ ? ) Wait compute: 2008^2=4032064, /2=2016032, then ( n=(2016032+2008)/3 = 2018040/3=672680 ). That gives huge n but valid. So infinite? No, finite divisors. So finite solutions. Find all functions ( f: \mathbbR \to \mathbbR
for a triangle with specific angles between the incentre, circumcentre, and vertices (
Angle between tangents at C and D = angle between (tangent at C, CA) + angle (CA, DA) + angle (DA, tangent at D) with signs. Using alternate segment: = angle CBA + angle CAD + angle DBA. But angle CAD is same as angle between CA and DA. And angle CBA + angle DBA = angle CBD. Notice that A,B,C,D concyclic? No, but A,B,C on circle1, A,B,D on circle2.