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.

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## Comments

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

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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.

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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.

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