MathDB

Problems(7)

Isogonal lines at the intersection of two circles

Source: RMO Delhi 2016, P3

10/11/2016
Two circles C1C_1 and C2C_2 intersect each other at points AA and BB. Their external common tangent (closer to BB) touches C1C_1 at PP and C2C_2 at QQ. Let CC be the reflection of BB in line PQPQ. Prove that CAP=BAQ\angle CAP=\angle BAQ.
geometry
number twice the square of sum of digits in decimal

Source: RMO Mumbai 2016, P3

10/11/2016
For any natural number nn, expressed in base 1010, let S(n)S(n) denote the sum of all digits of nn. Find all natural numbers nn such that n=2S(n)2n=2S(n)^2.
number theoryinequalities
All roots integers for a polynomial

Source: RMO Maharashtra and Goa 2016, P3

10/11/2016
Find all integers kk such that all roots of the following polynomial are also integers: f(x)=x3(k3)x211x+(4k8).f(x)=x^3-(k-3)x^2-11x+(4k-8).
number theoryalgebrapolynomial
Number Theory

Source: RMO 2016 Karnataka Region P3

10/16/2016
Let a,b,c,d,e,d,e,fa,b,c,d,e,d,e,f be positive integers such that ab<cd<ef\dfrac a b < \dfrac c d < \dfrac e f. Suppose afbe=1af-be=-1. Show that db+fd \geq b+f.
number theory
S(n) is sum of digits of n: n^3 = 8S(n)^3+6S(n)n+1

Source: RMO Hyderabad 2016 , P3 .

10/12/2016
For any natural number nn, expressed in base 1010, let S(n)S(n) denote the sum of all digits of nn. Find all positive integers nn such that n3=8S(n)3+6S(n)n+1n^3 = 8S(n)^3+6S(n)n+1.
number theory
RMO 2016 ,Q3

Source: Oct 23,2016

10/25/2016
The precent ages in years of two brothers AA and BB,and their father CC are three distinct positive integers a,ba ,b and cc respectively .Suppose b1a1\frac{b-1}{a-1} and b+1a+1\frac{b+1}{a+1} are two consecutive integers , and c1b1\frac{c-1}{b-1} and c+1b+1\frac{c+1}{b+1} are two consecutive integers . If a+b+c150a+b+c\le 150 , determine a,ba,b and cc.
number theory
2016 Chandigarh RMO (ad + bc) divides each of a, b, c,d

Source:

8/9/2019
a,b,c,da, b, c, d are integers such that ad+bcad + bc divides each of a,b,ca, b, c and dd. Prove that ad+bc=±1ad + bc =\pm 1
number theorydividesdivisor