22 SAMPLE PROBLEMS WITH SOLUTIONS FROM 555 GEOMETRY PROBLEMS

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1 22 SPL PROLS WITH SOLUTIOS FRO 555 GOTRY PROLS SOLUTIOS S O GOTRY I FIGURS Y. V. KOPY Stanislav hobanov Stanislav imitrov Lyuben Lichev

2 1 Problem 3.9. Let be a quadrilateral. Let J and I be the midpoints of the diagonals and, respectively. Let the perpendicular G to (G ) intersect the perpendicular H to (H ) at the point K. The perpendicular F to (F ) intersects the perpendicular to ( ) at the point L. Prove that KL JI. Solution. onsider the circle k 1 with diameter and the circle k 2 with diameter. Their centers are I and J, respectively. We have that KG.K = K.HK because the quadrilateral GH is cyclic. Therefore, the point K lies on the radical axis of k 1 and k 2. ut the same statement is also true for L because the quadrilateral F is cyclic, leading to L.LF = L.L. k 2 We know that the radical axis of two circles is always perpendicular to the line that connects their centers, and thus KL JI. F H J K k 1 L I G

3 2 Problems 3.11 and (Simson line) Let be a triangle with circumcircle k. n arbitrary point is chosen on the arc  of k which does not contain. The points, F and G lie on, and, respectively, and are chosen such that = F = G = ϕ. Prove that the points, F and G are collinear. Solution. Using the equal angles, it is easy to show that the quadrilaterals, F, F G and G are cyclic. ow we have F G = F + F G =180 + G = + G =180. F Therefore, the points, F and G are collinear. G ote. When ϕ = 90, the construction represents the classic Simson line. We proved the generalization of 3.11, which is 3.12.

4 3 Problem Let be a triangle. Let P be an arbitrary point on the smaller arc  of the circumcircle of. The projections of P onto and are X and Y, respectively. The points and are the midpoints of and XY, respectively. Prove that P = 90. Solution. Let K be the projection of P onto. Then the points X, Y and K lie on the Simson line of P (Problem 3.11). It suffices to prove that the quadrilateral P K is cyclic. X Y K P We show that P Y X P. We have P XY = P Y = P and XP Y = XY = P. ow the similarity follows and hence P K = P K because P and P are respective medians in these triangles. Therefore, the quadrilateral P K is cyclic.

5 4 Problem Let be a triangle with angle bisectors and, intersecting at the point I. Points L and K are chosen on the line, such that L and are symmetric with respect to, and such that and K are symmetric with respect to. Let = L K. Prove that I. Solution. The symmetry with respect to gives L =, and the symmetry with respect to gives K =. Therefore, LK is isosceles. Since I is the K-excenter of K, we deduce that KI is the angle bisector of LK. nalogously, LI is the angle bisector of KL. Therefore, I is the incenter of the isosceles LK, which implies that I. I L K

6 5 Problem Let be a triangle. Let its incircle touch the sides, and at the points F, and, respectively. Let its -excircle touch the lines, and at the points Q, P and, respectively. Let ( P Q) be an altitude in P Q, and let H ( F ) be an altitude in F. Prove that = H. Solution. It is enough to prove that sin P sin Q We have that = sin F H sin H. F = 90 β 2, F = 90 α 2. H F Q Hence, H = β 2 and HF = α 2. P Observe that P = β 2 and Q = α 2. by the law of sines we get Hence, P = 90 β 2, Q = 90 α 2 and sin P sin Q = P Q Q P = P Q = cot β 2 tg α 2 = H H HF H = HF H F = sin F H sin H.

7 6 Problem Let be a triangle. The points and are chosen on the sides and, respectively, so that is cyclic. Let = and = P. enote the midpoint of the segment by Q. Prove that the quadrilateral QP is cyclic. Solution. Without loss of generality, let >. enote = X. Then is between X and. enelaus Theorem, applied to and the points, and X, gives X X =.. Q P X eva s Theorem, applied to and the points P, and, gives P P =. = X X. ote that since X P > 1, the inequality > 1 holds, and so P is X P between Q and. We have that X.X = X.X, and so it suffices to show that X.X = XP.XQ. Let Q = a + b, QP = a, P = b and X = c. Then we have 2a + b 2a + 2b + c =, which implies ac = ab + b 2. b c quivalently, (b + c)(a + b + c) = c(2a + 2b + c). Therefore, XP.XQ = X.X.

8 7 Problem Let be a triangle with circumcenter O and = 60. Let 1 ( 1 ) and 1 ( 1 ) be altitudes in the triangle, intersecting at the point H. The line OH intersects the lines and at the points P and Q. Prove that P Q is equilateral. Solution. We have that O = H = 120, so the quadrilateral HO is cyclic. lso, O = 30, thus HO = 30. ote that H 1 = 60. Therefore, 1 HP = 30, which implies that QP = 60 and that P Q is equilateral. 1 P O H 1 Q

9 8 Problem Let be a parallelogram. Let be the midpoint of and let F be the foot of the perpendicular from to the line. Prove that F =. F P Solution. Let = P. Since = 2 and, we have that is a midsegment in P. Hence, P = and thus F is the median to the hypotenuse in the right-angled P F. We deduce that F = P =.

10 9 Problem Let be a square and let be a point on the segment. The square KF is constructed, such that it is external to. Let K = and K =. Prove that the points,, and F are collinear. F K Solution. We have = and = K. Hence, = K. Thus, K =. This implies that the quadrilateral is cyclic, which yields K = = 90. lso, F K = K = 90, so the quadrilateral F K is cyclic. Since K and K, the point is the orthocenter of K, which yields K = 90. onsequently, the quadrilateral K is cyclic. ow, = K = 45 = KF = 180 F implies that the points, and F are collinear. We have that is the radical axis of the circumcircles of K and. It suffices to show that also lies on this radical axis. We have = 45 = K = =. Thus, touches the circumcircle of. lso, = 45 = K and touches the circumcircle of K. Therefore, the point has equal powers with respect to these two circles.

11 10 Problem Let be a circumscribed quadrilateral with incircle k with center I. Let be also inscribed in a circle k 1. Let k touch,, and at the points,, P and Q, respectively. Prove that P Q. Solution. We have (P, Q) = = P + Q 2 P I + QI = 2 = ut we know that k 1 Q k I P + = 180. Hence, (P, Q) = 90.

12 11 Problem Let be a cyclic quadrilateral. Let,, P and Q be the midpoints of,, and, respectively. The points H 1, H 2, H 3 and H 4 are the feet of the perpendiculars from,, P and Q to,, and, respectively. Prove that the lines H 1, H 2, P H 3 and QH 4 are concurrent. Solution. idsegments give that P is parallel and equal to Q. Therefore, the quadrilateral P Q is a parallelogram. enote the circumcenter of by O. ow we have that OP, O, O and OQ. If H 1 P H 3 = X, then the quadrilateral P OX is a parallelogram. Thus, the midpoint of OX coincides with the midpoint of P, which coincides with the midpoint of Q. Hence, the quadrilateral OXQ is also a parallelogram. onsequently, X and QX, which gives QH 4 H 2 = X. Q H 2 H 3 X H 1 P O H 4

13 12 Problem Let k 1 and k 2 be circles that touch another circle k internally at the points and, respectively. Let k 1 k 2 = {, }. Prove that the intersection point L of the angle bisectors of and lies on. Solution. The angle bisector theorem, applied to and, yields that it suffices to show that =. Let be the intersection point of the tangent lines to k at and. Observe that lies on the radical axis of k 1 and k 2. Hence,. ote that and. Thus, L = = =.

14 13 Problem Let k, k 1 and k 2 be circles with radii r, r 1 and r 2, respectively. The circles k 1 and k 2 do not intersect each other and they touch k internally at the points and, respectively. The tangent lines and from to k 2 (, k 2 ) intersect k 1 at the points and, respectively. The tangent lines and F from to k 1 (, F k 1 ) intersect k 2 at the points Q and P, respectively. Let r be the radius of the circle that touches the segments and, and the arc of k 1. Let r be the radius of the circle that touches the segments P and Q, and the arc P Q of k 2. Prove that r = r. Solution. onsider the homothety centered at that sends k 1 to k. It sends the incircle of the curvilinear triangle to k 2 and hence r = r 1 r 2 r, so r = r 1.r 2 r. nalogously, considering the homothety centered at that sends k 2 to k, we get k 1 k F Q P k 2 r = r 1.r 2 r = r.

15 14 Problem (The butterfly theorem) Let be a quadrilateral inscribed in a circle with center O. The diagonals of intersect at the point. line perpendicular to O at intersects and at the points P and Q, respectively. Prove that P = Q. P Q O Solution. ote that = and =. Hence,. enote the midpoints of and by and, respectively. Then and are the corresponding medians of the similar triangles and. Thus, =. On the other hand, O. Hence, OP is cyclic and = P = P O. nalogously, OQ =. We have = and thus P O = QO. We obtained that the line O is both an altitude and an angle bisector in P OQ. Hence, it is a median as well.

16 15 Problem Let be a triangle. Let k(o) be the circumcircle of the triangle and let be the reflection of with respect to O. The tangent line to k at the point intersects at the point. Let O intersect and at the points F and P, respectively. Prove that F O = OP. Solution. Let F be a point, such that F. Let P be a point, such that P. Then clearly the quadrilateral F P is a parallelogram and the point O bisects its diagonals. Let F P =. We will prove that the point coincides with the point, which will lead us to P P and F F, so the desired statement will follow. F O P We have that = 2R, F = P = 90 β, P = F = 90 α and by the law of sines, F = 2R cos α sin γ, P = 2R cos β sin γ. Therefore, F = b F = 2R 2R cos β cos γ (sin β sin γ cos α) =, sin γ sin γ P = a P = 2R 2R cos α cos γ (sin α sin γ cos β) =. sin γ sin γ enelaus Theorem, applied to and the line P F, yields = 2R cos α cos γ sin γ 2R cos β sin γ 2R cos α sin γ 2R cos β cos γ sin γ = cos2 α cos 2 β.

17 16 On the other hand, we have that = = 90 + α. lso, = = 90 β. The law of sines, applied to and, yields = sin sin Hence,, as required. sin sin = cos2 α cos 2 β =. Problem The points,, and lie on a circle with diameter. The tangent lines to the circle at the points and intersect each other at the point F. Let =. Prove that F. Solution.. Let H = F enote = α and = β. Observe that = = 90 and = 90 α = F. nalogously, F = 90 β. We obtain = = 180 α β. It follows from the quadrilateral F that F = 2α + 2β, which means that 2(180 ) = F. Hence, lies on the circle centered at F with radius F. Thus, F = F. Therefore, H = F = F = 90 α, H = α and F. F H

18 17 Problem Let be a triangle with = 90. The tangent lines to the circumcircle of at and intersect each other at the point R. Let P and be the midpoints of and the arc Â, respectively. enote the second intersection point of P and the circumcircle of by Q. Prove that QR = 90. R P Q Solution. enote the midpoint of by. Hence, is the center of the circumcircle of, and the points, P and R are collinear. On the other hand, contains the midpoint of. Hence, and P, which means that P = 90. It suffices to show that the quadrilateral QR is cyclic. onsider the inversion I(, ). We will prove that the images of, Q and R are collinear. Observe that I() =, I(Q) = Q and I(R) = P. Since the points, Q and P are collinear, we conclude that the points, Q and R lie on a circle that contains the center of the inversion.

19 18 Problems Let k 1 and k 2 be circles that intersect each other at the points and. The tangent lines to k 1 and k 2 at and, respectively, intersect at the point O. Let k 1 and k 2 are points, such that O and O are the other tangent lines to k 1 and k 2, respectively. Prove that O = O. Solution. Let the mapping ϕ be a composition of inversion I(O, O.O) and symmetry with respect to the angle bisector of O. Then ϕ() =, ϕ() =, ϕ(k 1 ) = k 2 and ϕ(k 2 ) = k 1, because k 1 gets transformed into a circle that touches O at and that passes through. Thus, ϕ(o) = O, and so the lines O and O are symmetric with respect to the angle bisector of O, implying that O = O. k 1 k 2 O

20 19 Problem Let F be a regular hexagon. The points and are the midpoints of and F, respectively. Prove that is equilateral. Solution. It is clear that is equilateral. We have 2 = +. F +60 rotation of all vectors in the above equality gives 2 a = + F, where a is the image of the vector under a +60 rotation. Since =, we obtain 2 a = + F = 2, whence a =. Therefore, is equilateral. nother possible approach uses the fact that =.

21 20 Problem Let be an equilateral triangle and let l be an arbitrary line through the vertex. Let P and Q be the feet of the perpendiculars from and to l, respectively. If S is the midpoint of, prove that P QS is equilateral. Solution. Let and be the midpoints of and, respectively. onsidering the midsegments and the fact that Q and P are right-angled, we get that Q = = = S = S = = = P. oreover, P Q Q = Q = 180 2(120 P ) = 2 P 60 = 120 P. S Hence, SQ = Q = 240 P = P S. We deduce that P S = QS and P S = SQ. ut P SQ = S = = 60 and therefore P QS is equilateral.

22 21 Problem Let be a triangle. The isosceles triangles F (F = F ) and ( = ) are constructed externally, such that = F. The point lies inside of, such that = and =. Prove that F = and = F. Solution. The problem conditions imply that F. Then =. On the other hand, we have =, which implies that. F nalogously, F. Hence, F. This and the equality = imply that actually F =. Therefore, F = and = F.

23 22 Problem Let m, n and p be three rays with a common origin X. Let m and n. The line symmetric to with respect to n intersects p at the point. The line symmetric to with respect to p intersects m at the point. The line symmetric to with respect to m intersects n at the point. Finally, the line symmetric to with respect to n intersects p at the point F. Prove that the line F is symmetric to both F with respect to p and to with respect to m. Solution. Let the line symmetric to with respect to m intersect p at the point F. It suffices to show that F F. Recall that a point on an angle bisector is equidistant from the arms of the angle. We have dist(x, F ) = dist(x, ) m n X p F = dist(x, ) = dist(x, ) = dist(x, ) = dist(x, F ) Let the projections of X onto the lines F and F be H 1 and H 2, respectively. Then XH 2 = XH 1. We have H 1 F X = X + X XF = X + X X = X + X X = X + X XF = H 2 F X. Therefore, XH 1 F = XH 2 F, which means that XF = XF, or F F.

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