Tutor profile: Ali G.

There is point $$D$$ on edge $$AC$$ isosceles triangle $$ABC$$ with base $$BC$$. There is point $$K$$ on the smallest arc $$CD$$ of the circumcircle of triangle $$BCD$$. Ray $$CK$$ intersects line parallel to line $$BC$$ through $$A$$ at point $$T$$. Let $$M$$ be midpoint of segment $$DT$$. Prove that $$\angle AKT=\angle CAM$$ from all-Russian 2019 math olympiad and my solution:

Lemma: if we have $$\frac{sin(\alpha)}{sin(k-\alpha)} = \frac{sin(\beta)}{sin(k-\beta)}$$ then we have $$\alpha = \beta$$ or $$ k = 180$$ Main problem: Let $$X$$ be the intersection of the circle of triangle $$BDC$$ with $$AB$$. we have $$\angle XKT = \angle ABC = \angle ACB = \angle CAT$$ and we want to prove the equality of $$\angle AKT$$ and $$\angle CAM$$ and we have $$\angle TCB = \angle KXA = 180 - (\angle ATK)$$ thus we have $$sin(\angle ATK) = sin(\angle AXK)$$ . because of the lemma, we have : $$ \frac{sin(\angle AKT)}{sin(\angle AKZ)} = \frac {\frac{ AT . sin(\angle ATK)}{AK}}{ \frac{AX .sin(\angle AXK)}{AK}} = \frac{AT}{AX} = \frac{AT}{AD} = \frac{ sin(\angle TAM)}{sin( \angle DAM)}$$ thus we have $$\angle AKT = \angle CAM$$. DONE! (this is my own solution on AoPS with my account ALEEGH)

Prove that in a graph we have Eulerian path iff the degree of each vertex is even.

let $$G$$ be our graph and $$P$$ be our Eulerian path. In our path, when we visit a vertex, we enter the vertex once and exit from it once and we visit two new edges. thus for every entering, we have two edges, thus the degree of each vertex should be even.

let $$a,b,c$$ be three rational positive numbers. prove that: $$(a+b)(b+c)(c+a) \ge 8abc$$

because of $$AM-GM$$ we have: $$a+b \ge 2\sqrt{ab}$$ thus we have: $$(a+b)(b+c)(c+a) \ge (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ca}) = 8abc $$