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YIU : Problems in Elementary Geometry 1 American Mathematical Monthly Geometry Problems 1894 – Vis171. (Marcus Baker) In a traingle ABC, the center of the circumscribed circle is O, the center of the inscribed circle is I, and the orthocenter is H. Knowing the sides of the triangle OIH, determine the sides of triangle ABC. R − 2 Solution by W.P. Casey: Let N be the nine-point center. From IN = 2 r and OI = R(R − 2r), R and r can be determined. The circumcircle and the incircle, and the nine-point circle, can all be constructed. Then, Casey wrote, “[i]t only remains to find a point C in the circumcircle of ABC so that the tangent CA CA to circle I may be bisected by the circle N in the point S, which is easily done”. [sic] The construction of ABC from OIH cannot be effected by ruler and compass in general. Casey continued to derive the cubic equation with roots cos α, cos β,cosγ, and coefficients in terms of R, r,and := OH,namely, r 4(R + r)2 − (2 +3R2) R2 − 2 x3 − (1 + )x2 + x − =0. R 8R2 8R2 Given triangle OIH,letN be the midpoint of OH. Construct the circle through N tangent to OI at O.ExtendIN to intersect this circle again at M. The diameter of the circumcircle is equal to the length of IM. From this, the circumcircle, the nine-point circle, and the incircle can be constructed. Now, it remains to select a point X on the nine point circle so that the perpendicular to OX is tangent to the incircle. This is in general not constructible. √ √ √ √ Here is an example: Let OI =2 2α, OH = 42α,andIH = 15α,whereα = 17. Then s − a, s − b, s − c are the roots of the equation x3 − 34x2 +172x − 2 · 172 =0. YIU : Problems in Elementary Geometry 2 The roots of this equation are inconstructible. Vis186. (Marcus Baker) (Malfatti circles) Inscribe in any plane triangle three circles each tangent to two sides and the other two circles. Three solutions were published. The second one, by U. Jesse Knisely, contains a construction: Thethirdsolution,byE.B.Seitz,1 gives the radii of the Malfatti circles as (1 + tan β )(1 + tan γ ) 4 4 · r x = α , 1+tan 4 2 and analogous expressions for the other two. Vis52. (James McLaughlin) (Inverse Malfatti problem) Three circles, radii p, q, r,are drawn in a triangle, each circle touching the other two and two sides of the triangle. Find the sides of the triangle. Solution by E.B.Seitz: write σ = p + q + r. √ √ √ √ √ √ √ √ √ pσ +(p( p + q + r) − pqr) σ − pqr( q + r − p) a = √ √ √ √ √ √ √ √ (q + r)2. (p q + q p + pqσ − r σ)(p r + r p + prσ − q σ) Vis289. (Christine Ladd) If R is the radius of the circumscribed circle of a triangle ABC, r the radius of the inscribed circle, p the radius of the circle inscribed in the orthic triangle, I the center of the inscribed circle of the triangle ABC,andQ the center of the circle inscribed 1The very first issue of the Monthly contains a biography of Seitz, written by Finkel. YIU : Problems in Elementary Geometry 3 in the triangle formed by joining the midpoints of the sides of ABC, show that 1 QI2 = . 2(7Rp − 6Rr +3r2 +2R2) This certainly is not right! The correct formula should be 1 QI2 = (2Rp − 6Rr +3r2 +2R2). 2 Q is the Spieker center X10 and 1 QI2 = (−a3 +2a2(b + c) − 9abc). 8s IG 2 Also, IX10 = 3 .Now, − 3 2 − − 3 − 2 − 2 ( a +2a (b + c) 9abc)= s1 +5s1s2 18s3 =2s(5r 16Rr + s ). 2 1 2 − 2 Therefore, QI = 4 (5r 16Rr + s ). YZ Now, for the orthic triangle, sin α =2R cos α. This means that YZ = R sin 2α.Thesemi- perimeter of the orthic triangle is therefore 1 abc 4·R s = R(sin 2α +sin2β +sin2γ)=2R sin α sin β sin γ = = = . 2 4R2 4R2 R Also, =2cosα cos β cos γ. YIU : Problems in Elementary Geometry 4 The inradius of the orthic triangle is p = =2R cos α cos β cos γ. s Now, (b2 + c2 − a2)(c2 + a2 − b2)(a2 + b2 − c2) cos α cos β cos γ = 8a2b2c2 − 6 4 − 2 2 − 3 − 2 ( s1 +6s1s2 8s1s2 8s1s3 +16s1s2s3 8s3 = 2 8s3 32r2s2[s2 − (2R + r)2] = 8(4Rrs)2 s2 − (2R + r)2 = . 8R2 s2−(2R+r)2 Therefore, p = 4R . s2 =(2R + r)2 +4Rp. Finally, 1 QI2 = (5r2 − 16Rr + s2) 4 1 = [5r2 − 16Rr +(2R + r)2 +4Rp] 4 1 = [6r2 − 12Rr +4R2 +4Rp] 4 1 = (2Rp − 6Rr +3r2 +2R2). 2 VIS 274. If the midpoints of three sides of a triangle are joined with the opposite vertices and R1,...,R6 are the radii of the circles circumscribed about the 6 triangles so formed, and r1, r2,...,r6 the radii of the circles inscribed in these triangles. prove that R1R3R5 = R2R4R6 and 1 1 1 1 1 1 + + = + + . r1 r3 r5 r2 r4 r6 VIS 282 (W.P.Casey) Let X, Y , Z be the traces of the incenter. Construct circles through I and with centers at X, Y , Z. The sum of the reciprocals of the radii of all the circles touching these circles is equal to four times the reciprocal of the inradius. YIU : Problems in Elementary Geometry 5 G1.941.S941. (B.F.Finkel) Show that the bisectors of the angles formed by producing the sides of an inscribed quadrilateral intersect each other at right angles. G2.941.S942;943. (From Bowser’s Trigonometry) Show that π 2 · 4 · 6 · 8 · 10 ··· 2 = , 2 1 · 3 · 5 · 7 · 9 ··· Wallis’ expression for π. P.H. Philbrick pointed out that this expression was not correct. It should be π 2 · 4 · 6 · 8 · 10 ···(2n) 2 1 = · . 2 1 · 3 · 5 · 7 · 9 ···(2n − 1) 2n +1 G3.941. (From Todhunter’s Trigonometry)IfA be the area of the circle inscribed in a triangle, A1, A2, A3 the ares of the escribed circles, show that 1 1 1 1 √ = √ + √ + √ . A A1 A2 A3 G4.941.S943. (From Todhunter’s Trigonometry) Three circles whose radii are a, b,andc touch each other externally; prove that the tangents at the points of contact meet in a point whose distance from any one of them is abc . a + b + c YIU : Problems in Elementary Geometry 6 G5.941.S943. (Adolf Bailoff) If from a variable point in the base of an isosceles triangle, perpendiculars are drawn to the sides, the sum of the perpendiculars is constant and equal to the perpendicular let fall from either extremity of the base to the opposite side. G6.941.S94:159–160. (Earl D. West) Having given the sides 6, 4, 5, and 3 respectively of a trapezium, inscribe in a circle, to find the diameter of the circle. [Note: trapezium means quadrilateral]. G7.941.S94:160. (William Hoover) Through each point of the straight line x = my + h is drawn a chord of the parabola y2 =4ax, which is bisected in the point. Prove that this chord touches the parabola (y − 2mn)2 =8a(x − h). G8.941.S944. (Adolf Bailoff) If the two exterior angles at the base of a triangle are equal, the triangle is isosceles. G9.941.S94:160–161. (J.C.Gregg) Two circles intersect in A and B. Through A two lines CAE and DAF are drawn, each passing through a center and terminated by the circumferences. Show that CA · AE = DA · AF . (Euclid) G10.941.S94:161. (Eric Doolittle) If MN be any plane, and A and B any point without the plane, to find a point P , in the plane, such that AP + PB shall be a minimum. G11.941. (Lecta Miller) A gentleman’s residence is at the center of his circular farm containing a = 900 acres. He gives to each of his m = 7 children an equal circular farm as large as can be made within the original farm; and he retains as large a circular farm of which his residence is the center, as can be made after the distribution. Required the area of the farms made. YIU : Problems in Elementary Geometry 7 G12.S94:198–199. (J.F.W.Scheffer) Let OA and OB represent two variable conjugate semi - diameters of the ellipse x2 y2 + =1. a2 b2 On the chord AB as a side describe an equilateral triangle ABC. Find the locus of C. G13.941. (Henry Heaton) Through two given points to pass four spherical surfaces tangent to two given spheres. G14.S94:232–233;268–269. (Henry Heaton) Through a given point to draw four circles tangent to two given circles. G15.S94:233–234. (Issac L. Beverage) A man starts from the center of a circular 10 acre field and walks due north a certain distance, then turns and walks south - west till he comes to the circumference, walking altogether 40 rods. How far did he walk before making the turn? G16.941. (H.C.Whitaker) Three lights of intensities 2, 4, and 5 are placed respectively at points the coordinates of which are (0, 3), (4, 5) and (9, 0). Find a point in the plane of the lights equally illuminated by all of them. G17.942:S94:269. (Robert J. Aley) Draw a circle bisecting the circumference of three given circles. G18.942.S94:271. (Henry Heaton) Through two given points to draw two circles tangnet to a given circle. G19.942:S94:315.
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  • (Theorem 1) Is Proved

    (Theorem 1) Is Proved

    Rad HAZU Volume 503 (2009), 41–54 ABOUT ONE RELATION CONCERNING TWO CIRCLES MIRKO RADIC´ AND ZORAN KALIMAN Abstract. This article can be considered as an appendix to the article [1]. Here the article [1] is extended to the cases when one circle is outside of the other and when circles are intersecting. 1. Preliminaries In [1] the following theorem (Theorem 1) is proved: Let C1 and C2 be any given two circles such that C1 is inside of the C2 and let A1 , A2 , A3 be any given three different points on C2 such that there are points T1 and T2 on C1 with properties |A1A2| = t1 +t2, |A2A3| = t2 +t3, (1) where t1 = |A1T1|, t2 = |T1A2|, t3 = |T2A3|.Then 2rR |A A | =(t +t ) , (2) 1 3 1 3 R2 − d2 where r = radius of C1 , R = radius of C2 , d = |IO|, I is the center of C1 , O is center of C2 .(SeeFigure1.) In short about the proof of this theorem. First the following lemma is proved. It t1 is given then t2 can be calculated using the expression √ t (R2 − d2) ± D (t ) = 1 1 (3a) 2 1,2 2 + 2 r t1 where = 2( 2 − 2)2 +( 2 + 2) 2 2 − 2 2 − ( 2 + 2 − 2)2 . D1 t1 R d r t1 4R d r t1 R d r (3b) The values (t2)1,2 given by (3) are solutions of the equation ( 2 + 2) 2 − ( 2 − 2)+ 2 2 − 2 2 +( 2 + 2 − 2)2 = . r t1 t2 2t1t2 R d r t1 4R d R d r 0 (4) Mathematics subject classification (2000): 51M04.