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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 $$ABC$$ whose lengths of heights $$h_a$$ and $$h_b$$ (from A and B, respectively) and length of median $$m_a$$ (from A) are given.

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

(a) Find the locus of all midpoints of segments $$XY$$, where $$X$$ is any point on segment $$AC$$ and $$Y$$ any point on segment $$B'D'$$.

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

6. (BUL) An isosceles trapezoid with bases $$a$$ and $$b$$ and height $$h$$ 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 $$90^{\circ}$$.

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

(c) Under what conditions for $$a, b$$, and $$h$$can the point $$P$$ 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 $$V_1$$ be the volume of the cone and $$V_2$$ the volume of the cylinder.

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

(b) Find the smallest $$k$$ for which $$V_1 = k V_2$$, 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
11 months, 1 week ago

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My solution to Q4:

Construction: Draw segment $$AM$$ with length $$M_a$$. Now, draw a circle centred at $$A$$ with radius $$h_a$$. Draw one of the tangents of this circle passing through $$M$$. Now construct a circle with radius $$\dfrac {h_b}{2}$$ centred at $$M$$. Draw one of the tangents of this second circle passing through $$A$$. The intersection of these two tangents is $$C$$. Draw a line parallel to the second tangent with a distance of $$h_b$$ between them. The intersection of this line and the firs tangent is $$B$$. Thus, we have the 3 points $$A$$, $$B$$ and $$C$$ to construct triangle $$ABC$$.

Explanation: In a triangle $$ABC$$, points $$M$$ and $$N$$ are the midpoints of sides $$BC$$ and $$AC$$ respectively. We have that $$\Delta MNC \sim \Delta ABC$$ and as a result point $$M$$ is a distance of $$\dfrac {h_b}{2}$$ away from $$AC$$. The rest is self-explanatory. · 11 months ago

Sketch for a brutal solution for $$4$$.

From $$h_a,m_a$$ we can construct the vertex $$A$$, its foot of perpendicular on and midpoint of $$BC$$, which are denoted $$D,M$$ respectively. From $$h_a,h_b$$ we know the ratio of $$AC,BC$$ and $$MC,AC$$; specifically $$\frac {MC}{AC}=\frac {h_b}{2h_a}$$. Thus we can construct $$C$$ with the use of Apollonius circle, and $$B$$ consequently follows. · 11 months, 1 week ago