MathDB

2

Part of 2011 Belarus Team Selection Test

Problems(6)

AP x BH = BQ x AH, bisector, altitude, projections related

Source: 2011 Belarus TST 1.2

6/14/2020
Points LL and HH are marked on the sides ABAB of an acute-angled triangle ABC so that CLCL is a bisector and CHCH is an altitude. Let P,QP,Q be the feet of the perpendiculars from LL to ACAC and BCBC respectively. Prove that AP⋅BH=BQ⋅AHAP \cdot BH = BQ \cdot AH.
I. Gorodnin
geometryangle bisectoraltituderatio
<ACX=<BCY if AX x BX / CX^2 = AY x BY / CY^2

Source: 2011 Belarus TST 2.2

6/14/2020
Two different points X,YX,Y are marked on the side ABAB of a triangle ABCABC so that AX⋅BXCX2=AY⋅BYCY2\frac{AX \cdot BX}{CX^2}=\frac{AY \cdot BY}{CY^2} . Prove that ∠ACX=∠BCY\angle ACX=\angle BCY.
I.Zhuk
equal anglesgeometry
AB^2+DC^2=AC^2+BD^2, orthocenter, ext.angle bisector, circumcircle related

Source: 2011 Belarus TST 3.2

6/14/2020
The external angle bisector of the angle AA of an acute-angled triangle ABCABC meets the circumcircle of △ABC\vartriangle ABC at point TT. The perpendicular from the orthocenter HH of △ABC\vartriangle ABC to the line TATA meets the line BCBC at point PP. The line TPTP meets the circumcircce of △ABC\vartriangle ABC at point DD. Prove that AB2+DC2=AC2+BD2AB^2+DC^2=AC^2+BD^2
A. Voidelevich
geometrycircumcircleangle bisectororthocenter
m^2 +25 \vdots (2011^x-1007^y) where m,x,y in N

Source: 2011 Belarus TST 6.2

11/7/2020
Do they exist natural numbers m,x,ym,x,y such that m2+25⋮(2011x−1007y)?m^2 +25 \vdots (2011^x-1007^y) ? S. Finskii
number theorydividesdivisible
modified version of IMO 2010 SL G3 max { X_iX_{i+1}/P_iP_{i+1}) >=1

Source: 2011 Belarus TST 8.2

11/8/2020
Let A1A2…AnA_1A_2 \ldots A_n be a convex polygon. Point PP inside this polygon is chosen so that its projections P1,…,PnP_1, \ldots , P_n onto lines A1A2,…,AnA1A_1A_2, \ldots , A_nA_1 respectively lie on the sides of the polygon. Prove that for points X1,…,XnX_1, \ldots , X_n on sides A1A2,…,AnA1A_1A_2, \ldots , A_nA_1 respectively, max⁡{X1X2P1P2,…,XnX1PnP1}≥1.\max \left\{ \frac{X_1X_2}{P_1P_2}, \ldots, \frac{X_nX_1}{P_nP_1} \right\} \geq 1. if a) X1,…,XnX_1, \ldots , X_n are the midpoints of the corressponding sides, b) X1,…,XnX_1, \ldots , X_n are the feet of the corressponding altitudes, c) X1,…,XnX_1, \ldots , X_n are arbitrary points on the corressponding lines.
Modified version of [url=https://artofproblemsolving.com/community/c6h418634p2361975]IMO 2010 SL G3 (it was question c)
geometrygeometric inequality
sum a^3 bc/(b+c) >= 1/6 (sum ab)^2 if sum a/(b+c)=1+1/6 (sum a/c) for a,b,c>0

Source: 2011 Belarus TST 5.2

11/7/2020
Positive real a,b,ca,b,c satisfy the condition ab+c+ba+c+ca+b=1+16(ac+ba+cb)\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1+\frac{1}{6}\left( \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \right) Prove that a3bcb+c+b3caa+c+c3aba+b≥16(ab+bc+ca)2\frac{a^3bc}{b+c}+\frac{b^3ca}{a+c}+\frac{c^3ab}{a+b}\ge \frac{1}{6}(ab+bc+ca)^2 I.Voronovich
algebrainequalities