IMO 1960

IMO 1960 — 7/7 solved, 6 verified.

7 items

IMO 1960

Official IMO 1960 problems  ·  7/7 solved, 6 verified.

# Status Time
1 solved 27m41s
2 ✓ verified 7m41s
3 ✓ verified 17m36s
4 ✓ verified 33m08s
5 ✓ verified 4m30s
6 ✓ verified 3m00s
7 ✓ verified 5m59s

Problem 1   solved · 27m41s · Solution →

Find all three-digit numbers for which one obtains, when dividing the number by 11, the sum of the squares of the digits of the initial number.

Problem 2   ✓ verified · 7m41s · Solution →

For which real numbers $x$ does the following inequality hold: $$\frac{4x^2}{\left(1 - \sqrt{1+2x}\right)^2} < 2x + 9\,?$$

Problem 3   ✓ verified · 17m36s · Solution →

A right-angled triangle $ABC$ is given for which the hypotenuse $BC$ has length $a$ and is divided into $n$ equal segments ($n$ odd). Let $\alpha$ be the angle subtended at $A$ by the segment containing the midpoint of $BC$. If $h$ denotes the altitude from $A$, prove that $$\tan\alpha = \frac{4nh}{(n^2-1)a}.$$

Problem 4   ✓ verified · 33m08s · Solution →

Construct a triangle $ABC$ given the lengths of the altitudes $h_a$ and $h_b$ (from $A$ and $B$ respectively) and the length of the median $m_a$ from $A$.

Problem 5   ✓ verified · 4m30s · Solution →

Consider the cube $ABCDA'B'C'D'$ (with face $ABCD$ directly above face $A'B'C'D'$).

a) Find the locus of the midpoints of the segments $XY$, where $X$ is any point of $AC$ and $Y$ is any point of $B'D'$;

b) Find the locus of points $Z$ which lie on the segment $XY$ of part a) with $ZY = 2XZ$.

Problem 6   ✓ verified · 3m00s · Solution →

Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ be the volume of the cylinder.

a) Prove that $V_1 \neq V_2$;

b) Find the smallest number $k$ for which $V_1 = kV_2$; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.

Problem 7   ✓ verified · 5m59s · Solution →

A regular cone is inscribed in a sphere. Around the sphere a cylinder is circumscribed so that its base lies in the same plane as the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ that of the cylinder.

(a) Prove that $V_1 = V_2$ is impossible.

(b) Find the smallest $k$ for which $V_1 = k V_2$, and in this case construct the half-angle at the vertex of the cone.