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**Extra resources for 44th International Mathematical Olympiad: Short-listed problems and solutions**

**Sample text**

A1 j] from [b2m−1 · · · b1 ]. Hence we have |A0 | = 10|B0 |, and so f (2m) = 10f (2m − 1). 31 Geometry G1. Let ABCD be a cyclic quadrilateral. Let P , Q, R be the feet of the perpendiculars from D to the lines BC, CA, AB, respectively. Show that P Q = QR if and only if the bisectors of ∠ABC and ∠ADC are concurrent with AC. Solution 1. A R D Q B C P It is well-known that P , Q, R are collinear (Simson’s theorem). Moreover, since ∠DP C and ∠DQC are right angles, the points D, P , Q, C are concyclic and so ∠DCA = ∠DP Q = ∠DP R.

1 + 1 + c2 Suppose that O1 (0, β) is the centre of the circumcircle Γ1 of the triangle AIB. Since (β − α)2 = O1 I 2 = O1 A2 = 1 + β 2 , we have β = −1/c and so Γ1 : x2 + (y + 1/c)2 = 1 + (1/c)2 . Let P (p, q). Since D(p − q/c, 0), E(p + q/c, 0), F (q/c − 1, q), G(−q/c + 1, q), it follows that the equations of the lines DF and EG are y= 2q c q q x− p− c −p−1 and y = − 2qc q q x− p+ c −p+1 , respectively. Therefore the intersection Q of these lines is (q − c)p/(2q − c), q 2 /(2q − c) . Let O2 (0, γ) be the circumcentre of the triangle ABC.

Further assume that a has n + 1 digits, n ≥ 0. Let s be the last digit of a and f the first digit of c. Since (∗ · · · ∗ s)2 = a2 = d = ∗ · · · ∗ f and (s ∗ · · · ∗)2 = b2 = c = f ∗ · · · ∗, where the stars represent digits that are unimportant at the moment, f is both the last digit of the square of a number that ends in s and the first digit of the square of a number that starts in s. The square a2 = d must have either 2n + 1 or 2n + 2 digits. If s = 0, then n = 0, b has n digits, its square c has at most 2n digits, and so does d, a contradiction.

### 44th International Mathematical Olympiad: Short-listed problems and solutions

by David

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