International Mathematical Olympiad 60', Second Day

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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
3 years, 6 months ago

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

Sualeh Asif - 3 years, 6 months ago

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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.

Xuming Liang - 3 years, 6 months 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.

Sharky Kesa - 3 years, 6 months ago

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