International Mathematical Olympiad 60', Second Day

The goal of this set of notes is to improve our problem solving and proof writing skills. You are encouraged to submit a solution to any of these problems, and join in the discussion in #imo-discussion on Saturday 24 at 9:00 pm IST ,8 30 PDT. For more details, see IMO Problems Discussion Group.


Sorry Everyone!, I was really busy the last whole week. Here is the next set of problems from the 1960 IMO.

Second Day

4. (HUN) Construct a triangle ABCABC whose lengths of heights hah_a and hbh_b (from A and B, respectively) and length of median mam_a (from A) are given.

5. (CZS) A cube ABCDA'B'C'D' is given.

(a) Find the locus of all midpoints of segments XYXY, where XX is any point on segment ACAC and YY any point on segment B'D'.

(b) Find the locus of all points ZZ on segments XYXY such that ZY=2XZ.\vec{ZY} = 2 \vec{XZ}.

6. (BUL) An isosceles trapezoid with bases aa and bb and height hh is given.

(a) On the line of symmetry construct the point P such that both (nonbase) sides are seen from P with an angle of 9090^{\circ}.

(b) Find the distance of PP from one of the bases of the trapezoid.

(c) Under what conditions for a,ba, b, and hhcan the point PP be constructed (analyze all possible cases)?

7. (GDR) A sphere is inscribed in a regular cone. Around the sphere a cylinder is circumscribed so that its base is in the same plane as the base of the cone. Let V1V_1 be the volume of the cone and V2V_2 the volume of the cylinder.

(a) Prove that V1=V2V1 = V2 is impossible.

(b) Find the smallest kk for which V1=kV2V_1 = k V_2, and in this case construct the angle at the vertex of the cone.

This is part of the set International Mathematical Olympiads

Note by Sualeh Asif
4 years ago

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@Xuming Liang @Nihar Mahajan here is the whole load of Olympiad Geometry! Enjoy!

Sualeh Asif - 4 years ago

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Sketch for a brutal solution for 44.

From ha,mah_a,m_a we can construct the vertex AA, its foot of perpendicular on and midpoint of BCBC, which are denoted D,MD,M respectively. From ha,hbh_a,h_b we know the ratio of AC,BCAC,BC and MC,ACMC,AC; specifically MCAC=hb2ha\frac {MC}{AC}=\frac {h_b}{2h_a}. Thus we can construct CC with the use of Apollonius circle, and BB consequently follows.

Xuming Liang - 3 years, 12 months ago

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My solution to Q4:

Construction: Draw segment AMAM with length MaM_a. Now, draw a circle centred at AA with radius hah_a. Draw one of the tangents of this circle passing through MM. Now construct a circle with radius hb2\dfrac {h_b}{2} centred at MM. Draw one of the tangents of this second circle passing through AA. The intersection of these two tangents is CC. Draw a line parallel to the second tangent with a distance of hbh_b between them. The intersection of this line and the firs tangent is BB. Thus, we have the 3 points AA, BB and CC to construct triangle ABCABC.

Explanation: In a triangle ABCABC, points MM and NN are the midpoints of sides BCBC and ACAC respectively. We have that ΔMNCΔABC\Delta MNC \sim \Delta ABC and as a result point MM is a distance of hb2\dfrac {h_b}{2} away from ACAC. The rest is self-explanatory.

Sharky Kesa - 3 years, 12 months ago

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