Problems

There is a triangle with side lengths \(a\), \(b\) and \(c\). Can you form a triangle with side lengths \(\frac{a}{b}\), \(\frac{b}{c}\) and \(\frac{c}{a}\)? Does it depend on what \(a\), \(b\) and \(c\) are? Give a proof if it is always possible or never possible. Otherwise, construct examples to show the dependence on \(a\), \(b\) and \(c\).
Recall that a triangle can be drawn with side lengths \(x\), \(y\) and \(z\) if and only if \(x+y>z\), \(y+z>x\) and \(z+x>y\).

There is a triangle with side lengths \(a\), \(b\) and \(c\). Does there exist a triangle with side lengths \(|a-b|\), \(|b-c|\) and \(|c-a|\)? Does it depend on what \(a\), \(b\) and \(c\) are?
Recall that a triangle can be formed with side lengths \(x\), \(y\) and \(z\) if and only if all the inequalities \(x+y>z\), \(y+z>x\) and \(z+x>y\) hold.

There is a triangle with side lenghts \(a\), \(b\) and \(c\). Does there exist a triangle with sides of lengths \(a^2+bc\), \(b^2+ca\) and \(c^2+ab\)? Does it depend on the values of \(a\), \(b\) and \(c\)?

Prove the triangle inequality: in any triangle \(ABC\) the side \(AB < AC+ BC\).

In certain kingdom there are a lot of cities, it is known that all the distances between the cities are distinct. One morning one plane flew out of each city to the nearest city. Could it happen that in one city landed more than \(5\) planes?

Is it possible to draw \(K_5\) without intersecting edges on a Möbius band? Recall that \(K_5\) is the complete graph on \(5\) vertices. That is, \(5\) points with an edge between every pair of different points.

A circle is inscribed in a triangle (that is, the circle touches the sides of the triangle on the inside). Let the radius of the circle be \(r\) and the perimeter of the triangle be \(p\). Prove that the area of the triangle is \(\frac{pr}{2}\).

Suppose you only knew the formula of a triangle for right-angled triangles. That is, if a base with length \(b\) and a height \(h\) of a triangle meet at a right angle, you know that the area of the triangle is \(\frac{1}{2}bh\). Can you prove the usual area formula for a general triangle?

There is a pair of parallel lines. The point \(A\) and \(B\) lie on one of the lines. The point \(C\) and \(D\) lies on the other line. We can form triangles \(\triangle ABC\) and \(\triangle ABD\). Prove that the areas of triangles \(\triangle ABC\) and \(\triangle ABD\) are equal.