MathDB

P3

Part of 2023 Israel TST

Problems(5)

Function from integers to naturals

Source: 2023 Israel TST Test 1 P3

3/23/2023
Find all functions f:Z→Z>0f:\mathbb{Z}\to \mathbb{Z}_{>0} for which f(x+f(y))2+f(y+f(x))2=f(f(x)+f(y))2+1f(x+f(y))^2+f(y+f(x))^2=f(f(x)+f(y))^2+1 holds for any x,y∈Zx,y\in \mathbb{Z}.
TSTfunctional equationalgebrafunction
Erecting Trapezoids from triangle

Source: 2023 Israel TST Test 3 P3

3/23/2023
Let ABCABC be a fixed triangle. Three similar (by point order) isosceles trapezoids are built on its sides: ABXY,BCZW,CAUVABXY, BCZW, CAUV, such that the sides of the triangle are bases of the respective trapezoids. The circumcircles of triangles XZU,YWVXZU, YWV meet at two points P,QP, Q. Prove that the line PQPQ passes through a fixed point independent of the choice of trapezoids.
geometrytrapezoidTSTcircumcircle
Complementary or equal angles

Source: 2023 Israel TST Test 2 P3

3/23/2023
In triangle ABCABC the orthocenter is HH and the foot of the altitude from AA is DD. Point PP satisfies AP=HPAP=HP, and the line PAPA is tangent to (ABC)(ABC). Line PDPD intersects lines AB,ACAB, AC at points X,YX,Y respectively.
Prove that ∠YHX=∠BAC\angle YHX = \angle BAC or ∠YHX+∠BAC=180∘\angle YHX+\angle BAC= 180^\circ.
TSTgeometry
Perfect polynomials

Source: 2023 Israel TST Test 5 P3

3/23/2023
Given a polynomial PP and a positive integer kk, we denote the kk-fold composition of PP by P∘kP^{\circ k}. A polynomial PP with real coefficients is called perfect if for each integer nn there is a positive integer kk so that P∘k(n)P^{\circ k}(n) is an integer. Is it true that for each perfect polynomial PP, there exists a positive mm so that for each integer nn there is 0<k≤m0<k\leq m for which P∘k(n)P^{\circ k}(n) is an integer?
TSTnumber theoryPolynomialscompositionalgebra
Projections to $OI$

Source: 2023 Israel TST Test 7 P3

5/9/2023
Let ABCABC be an acute-angled triangle with circumcenter OO and incenter II. The midpoint of arc BCBC of the circumcircle of ABCABC not containing AA is denoted SS. Points E,FE, F were chosen on line OIOI for which BEBE and CFCF are both perpendicular to OIOI. Point XX was chosen so that XE⊥ACXE\perp AC and XF⊥ABXF\perp AB. Point YY was chosen so that YE⊥SCYE\perp SC and YF⊥SBYF\perp SB. DD was chosen on BCBC so that DI⊥BCDI\perp BC. Prove that XX, YY, and DD are collinear.
TSTgeometrytriangle centerscircumcircleincenter