MathDB

1

Part of 2003 China Team Selection Test

Problems(10)

constructed triangle is right-angled

Source: China Team Selection Test 2003, Day 1, Problem 1

10/13/2005
ABCABC is an acute-angled triangle. Let DD be the point on BCBC such that ADAD is the bisector of ∠A\angle A. Let E,FE, F be the feet of perpendiculars from DD to AC,ABAC,AB respectively. Suppose the lines BEBE and CFCF meet at HH. The circumcircle of triangle AFHAFH meets BEBE at GG (apart from HH). Prove that the triangle constructed from BGBG, GEGE and BFBF is right-angled.
geometrycircumcircletrigonometryangle bisectorgeometry solved
f(n+1) >= f(n)

Source: China Team Selection Test 2003, Day 2, Problem 1

10/13/2005
Find all functions f:Z+→Rf: \mathbb{Z}^+\to \mathbb{R}, which satisfies f(n+1)≥f(n)f(n+1)\geq f(n) for all n≥1n\geq 1 and f(mn)=f(m)f(n)f(mn)=f(m)f(n) for all (m,n)=1(m,n)=1.
functioninductionlogarithmsalgebra unsolvedalgebra
x+y+z=xyz

Source: China TST 2003

6/29/2006
xx, yy and zz are positive reals such that x+y+z=xyzx+y+z=xyz. Find the minimum value of: x7(yz−1)+y7(zx−1)+z7(xy−1) x^7(yz-1)+y^7(zx-1)+z^7(xy-1)
inequalitiesinequalities unsolved
circumscribed quadrilateral [OA*OC + OB*OD = sqrt(abcd)]

Source: Chinese TST 2003

9/20/2005
Let ABCD ABCD be a quadrilateral which has an incircle centered at O O. Prove that OA\cdot OC\plus{}OB\cdot OD\equal{}\sqrt{AB\cdot BC\cdot CD\cdot DA}
geometrytrigonometrycalculusgeometry proposed
Find number of elements

Source: China TST 2003

6/29/2006
mm and nn are positive integers. Set A={1,2,⋯ ,n}A=\{ 1, 2, \cdots, n \}. Let set Bnm={(a1,a2⋯ ,am)∣ai∈A,i=1,2,⋯ ,m}B_{n}^{m}=\{ (a_1, a_2 \cdots, a_m) \mid a_i \in A, i= 1, 2, \cdots, m \} satisfying: (1) ∣ai−ai+1∣≠n−1|a_i - a_{i+1}| \neq n-1, i=1,2,⋯ ,m−1i=1,2, \cdots, m-1; and (2) at least three of a1,a2,⋯ ,ama_1, a_2, \cdots, a_m (m≥3m \geq 3) are pairwise distince. Find ∣Bnm∣|B_n^m| and ∣B63∣|B_6^3|.
algebra unsolvedalgebra
Find the side lengths of triangle

Source: China TST 2003

6/29/2006
In triangle ABCABC, AB>BC>CAAB > BC > CA, AB=6AB=6, ∠B−∠C=90o\angle{B}-\angle{C}=90^o. The incircle touches BCBC at EE and EFEF is a diameter of the incircle. Radical AFAF intersect BCBC at DD. DEDE equals to the circumradius of △ABC\triangle{ABC}. Find BCBC and ACAC.
geometrycircumcirclegeometry proposed
Colour the points

Source: China TST 2003

6/29/2006
Let SS be the set of points inside and on the boarder of a regular haxagon with side length 1. Find the least constant rr, such that there exists one way to colour all the points in SS with three colous so that the distance between any two points with same colour is less than rr.
inequalitiesreal analysiscombinatorics unsolvedcombinatorics
Sum a_k =<n

Source: China TST 2003

6/29/2006
Let g(x)=∑k=1nakcos⁡kxg(x)= \sum_{k=1}^{n} a_k \cos{kx}, a1,a2,⋯ ,an,x∈Ra_1,a_2, \cdots, a_n, x \in R. If g(x)≥−1g(x) \geq -1 holds for every x∈Rx \in R, prove that ∑k=1nak≤n\sum_{k=1}^{n}a_k \leq n.
trigonometryinequalities unsolvedinequalities
Area covered by circles

Source: China TST 2003

6/29/2006
There are nn(n≥3n\geq 3) circles in the plane, all with radius 11. In among any three circles, at least two have common point(s), then the total area covered by these nn circles is less than 3535.
geometrygeometry unsolved
Equal segments

Source: China TST 2003

6/29/2006
Triangle ABCABC is inscribed in circle OO. Tangent PDPD is drawn from AA, DD is on ray BCBC, PP is on ray DADA. Line PUPU (U∈BDU \in BD) intersects circle OO at QQ, TT, and intersect ABAB and ACAC at RR and SS respectively. Prove that if QR=STQR=ST, then PQ=UTPQ=UT.
geometrycircumcircletrapezoidperpendicular bisectorgeometry unsolved