MathDB

4

Part of 2015 Czech-Polish-Slovak Junior Match

Problems(2)

P is incenter of the CDE wanted, starting with a right triangle

Source: Czech-Polish-Slovak Junior Match 2015, Individual p4 CPSJ

3/15/2020
Let ABCABC ne a right triangle with ∠ACB=90o\angle ACB=90^o. Let E,FE, F be respecitvely the midpoints of the BC,ACBC, AC and CDCD be it's altitude. Next, let PP be the intersection of the internal angle bisector from AA and the line EFEF. Prove that PP is the center of the circle inscribed in the triangle CDECDE .
geometryangle bisectorincenterright trianglemidpoints
a + b + (gcd (a, b))^ 2 = lcm (a, b) = 2 \cdot lcm(a -1, b)

Source: Czech-Polish-Slovak Junior Match 2015, Team p4 CPSJ

3/19/2020
Determine all such pairs pf positive integers (a,b)(a, b) such that a+b+(gcd(a,b))2=lcm(a,b)=2⋅lcm(a−1,b)a + b + (gcd (a, b))^ 2 = lcm (a, b) = 2 \cdot lcm(a -1, b), where lcm(a,b)lcm (a, b) denotes the smallest common multiple, and gcd(a,b)gcd (a, b) denotes the greatest common divisor of numbers a,ba, b.
LCMGCDnumber theory