Geometry

# Geometry Problems of RMO’17

Q.1. Let AOB be a given angle less than $180^{\circ}$ and let P be an interior point of the angular region determined by $\angle AOB$. Show, with proof, how to construct, using only ruler and compass, a line segment CD passing through P such that C lies on the way OA and D lies on the ray OB, and CP:PD=1:2.

Analysis: Draw a line parallel to OB through C. Let it intersect OP (extended) at K. $\triangle CPK$ and $\triangle DPO$ are similar which implies $OP : PK = DP : PC = 2:1.$

Construction: Choose K on extended OP such that OP : PK=2:1. Draw a line parallel to OB through K. It will intersect OA at C (required). Then CP will cut OB at D (required).

Q.5. Let $\Omega$ be a circle with a chord AB which is not a diameter. $\Gamma_{1}$ be a circle on one side of AB such that it is tangent to AB at C and internally tangent to $\Omega$ at D. Likewise, let $\Gamma_{2}$ be a circle on the other side of AB such that it is tangent to AB at E and internally tangent to $\Omega$ at F. Suppose the line DC intersects $\Omega$ at $X \neq D$ and the line FE intersects $\Omega$ at $Y \neq F$. Prove that XY is a diameter of $\Omega.$

Label the centres as P and Q, as shown in the figure. Let O be the centre of the main circle.

$\angle PCD=\angle PDC=\angle OXD$ gives $XO\parallel CP.$ Now, $CP\bot AB,$ so we get $XO\bot AB.$ Thus XO is nothing but the perpendicular bisector of AB. Similarly YO is also the perpendicular bisector of AB. Hence X, O, Y are collinear. Hence done.

## 4 thoughts on “Geometry Problems of RMO’17”

1. Shubhav Jain says:

Why:
Thus XO is nothing but the perpendicular bisector of AB.

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