MathDB

2019 Saint Petersburg MO Grade 11 P4

Source: 2019 Saint Petersburg

4/14/2019

A non-equilateral triangle

\triangle ABC

of perimeter

12

is inscribed in circle

\omega

.Points

P

and

Q

are arc midpoints of arcs

ABC

and

ACB

, respectively. Tangent to

\omega

at

A

intersects line

PQ

at

R

. It turns out that the midpoint of segment

AR

lies on line

BC

. Find the length of the segment

BC

.
(А. Кузнецов)

geometryperimeter

2 right angles, 2 centroids, 1 circumcircle given, equal angles wanted

Source: St. Petersburg 2019 10.4

5/1/2019

Given a convex quadrilateral

ABCD

. The medians of the triangle

ABC

intersect at point

M

, and the medians of the triangle

ACD

at point

N

. The circle, circumscibed around the triangle

ACM

, intersects the segment

BD

at the point

K

lying inside the triangle

AMB

. It is known that

\angle MAN = \angle ANC = 90^o

. Prove that

\angle AKD = \angle MKC

.

geometrycircumcircleCentroidequal angles

cards with divisors of 6^{100}

Source: St. Petersburg 2019 9.4

5/2/2019

Olya wrote fractions of the form

1 / n

on cards, where

n

is all possible divisors the numbers

6^{100}

(including the unit and the number itself). These cards she laid out in some order. After that, she wrote down the number on the first card, then the sum of the numbers on the first and second cards, then the sum of the numbers on the first three cards, etc., finally, the sum of the numbers on all the cards. Every amount Olya recorded on the board in the form of irreducible fraction. What is the least different denominators could be on the numbers on the board?

number theoryreciprocal sumSumDivisorsIrreducible

4

Problems(3)