Singapore International athematical Olympiad Training Problems 18 January 2003 1 Let be a point on the segment Squares D and EF are erected on the same side of with F lying on The circumcircles of D and EF meet at a second point N Prove that N is the intersection of the lines F and 2 Let D be the diameter of the incircle of a triangle where D is the point at which the incircle touches the side The extension of meets at K Prove that K = D 3 Tangents P and P are drawn from a point P outside a circle Γ line through P intersects at S and Γ at Q and R Prove that P S is the harmonic mean of P R and P Q 4 (IO 1981)Three circles of equal radius have a common point O and lie inside a given triangle Each circle touches a pair of sides of the triangle Prove that the incenter and the circumcenter of the triangle are collinear with the point O
1 Let be a point on the segment Squares D and EF are erected on the same side of with F lying on The circumcircles of D and EF meet at a second point N Prove that N is the intersection of the lines F and D N F E Let F intersect at N We wish to show that N = N s F is congruent to, we have N = 90 so that N lies on the circle with as diameter That is N lies on the circumcircle of D Similarly, N lies on the circumcircle of EF Thus N = N 2 Let D be the diameter of the incircle of a triangle where D is the point at which the incircle touches the side The extension of meets at K Prove that K = D K D O 2
onsider a homothety centered at carrying the incircle to the excircle The diameter D of the incircle is mapped to the diameter D of the excircle Since D is perpendicular to, D is also perpendicular to Therefore must be the point K That is the excircle touches at K Therefore, K = (a + b c)/2 = D 3 Tangents P and P are drawn from a point P outside a circle Γ line through P intersects at S and Γ at Q and R Prove that P S is the harmonic mean of P R and P Q Since P Q is similar to RP, we have P Q/P = Q/R lso P Q is similar to RP, we have P R/P = R/Q Dividing the second equation by the first equation and using the fact that P = P, we obtain P R/P Q = (R/Q) (R/Q) = (RS/S) (S/QS) = SR/SQ This shows that the ratio that S divides QR internally is the same as the ratio that P divides QR externally This determines the position of S on the segment QR Q S R P Thus SR = QR P R P R + P Q, SQ = QR P Q P R + P Q lso P S = P R + RS = P R + QR P R (P Q P R) P R = P R + = 2P R P Q P R + P Q P R + P Q P R + P Q That is P S is the harmonic mean of P R and P Q (Second by olin Tan) Let be the midpoint of QR Then to prove P S = 2P R P Q/(P R+P Q) is equivalent to prove that P S P = P R P Q Or equivalently, P S P = P 2, since P R P Q = P 2 Therefore, we have to show that P is tangent to the circumcircle of S Let O be the centre of Γ Then O,, P, are concyclic and OP is perpendicular to Hence, P = P O = S Therefore, P is tangent to the circumcircle of S 3
Q S R O P (Third ) pplying Stewart s Theorem to P, we have (S + S) P S 2 + (S + S) S S = P 2 S + P 2 S Since P = P, we may cancel the common factor (S + S), thus obtaining P S 2 + S S P 2 = 0 Since S S = QS SR = (P Q P S)(P S P R) = (P Q+P R) P S P S 2 P Q P R and P 2 = P Q P S, we have (P Q+P R) P S = 2P Q P R Thus, P S is the harmonic mean of P R and P Q 4 (IO 1981)Three circles of equal radius have a common point O and lie inside a given triangle Each circle touches a pair of sides of the triangle Prove that the incenter and the circumcenter of the triangle are collinear with the point O O I 4
Let,, be the centres of the circles inside s,, are angle bisectors, they meet at the incenter I of triangle I is also the incenter of the triangle The circles are of the same radii Thus and are of equal distance from so that is parallel to Similarly, is parallel to and is parallel to That is is similar to onsider a homothety centred at I sending to, to and to Thus the circumcentre O of is mapped to the circumcentre of under this homothety Therefore, I,, O are collinear 5