MathDB

5

Part of 2013 JBMO Shortlist

Problems(2)

Junior Geometric Inequality with <BAC=60

Source: JBMO Shortlist 2013, G5

6/11/2017
A circle passing through the midpoint MM of the side BCBC and the vertex AA of the triangle ABCABC intersects the segments ABAB and ACAC for the second time in the points PP and QQ, respectively. Prove that if ∠BAC=60∘\angle BAC=60^{\circ}, then AP+AQ+PQ<AB+AC+12BCAP+AQ+PQ<AB+AC+\frac{1}{2} BC.
geometrygeometric inequality
\frac{1}{x^2}+\frac{y}{xz}+\frac{1}{z^2}=\frac{1}{2013}

Source: JBMO Shortlist 2013 NT5

4/24/2019
Solve in positive integers: 1x2+yxz+1z2=12013\frac{1}{x^2}+\frac{y}{xz}+\frac{1}{z^2}=\frac{1}{2013} .
number theoryDiophantine equationpositive integers