Pure plane geometry problems for RMO training or practice

Similar Figures:

Definition:

Polygons which are equiangular and have their corresponding sides proportional are called similar.

If, in addition, their corresponding sides are parallel, they are said to be similarly situated or homothetic.

Theorem 1:

If O is any fixed point and ABCD…X any polygon, and if points $A^{'}, B^{'}, C^{'}, \ldots X^{'}$ are taken on OA, OB, OC, …OX (or those lines produced either way) such that $\frac{OA^{'}}{OA} = \frac{OB^{'}}{OB} = \ldots = \frac{OX^{'}}{OX}$, then the polygons $ABCD...X$ and $A^{'}B^{'}C^{'} \ldots X^{'}$ are homothetic.

Before we state the next theorem, some background is necessary.

If O is a fixed point and P is a variable point on a fixed curve S, and if $P^{'}$ is a point on OP such that $\frac{OP^{'}}{OP} = k$, a constant, then the locus of $P^{'}$ is a curve $S^{'}$, which is said to be homothetic to S; and $P, P^{'}$ are corresponding points.

O is called the centre of similitude of the two figures.

If $P$ and $P^{'}$ lie on the same side of O, the figures are said to be directly homothetic w.r.t.O, and O is called the external centre of similitude.

If $P$ and $P^{'}$ lie on the opposite sides of O, the figures are said to be inversely homothetic w.r.t. O, and O is called the internal centre of similitude.

If we join A and B in the first case, we say that the parallel lines $AB, A^{'}B^{'}$ are drawn in the same sense, and in the second case, in opposite senses.

Theorem 2:

Let A, B be the centres of any two circles of radii a, b; AB is divided externally at O and internally at $O_{1}$ in the ratio of the radii, that is, $\frac{AO}{BO} = \frac{AO_{1}}{O_{1}B} = \frac{a}{b}$ then the circles are directly homothetic w.r.t. O and inversely homothetic w.r.t. $O_{2}$, and corresponding points lie on the extremities of parallel radii.

Now, prove the above two hard core basic geometry facts. ! 🙂 🙂 🙂

Cheers,

Nalin Pithwa.

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