Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 9 Solutions

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1 Math 451: Euclidean and Non-Euclidean Geomety MWF 3pm, Gasson 04 Homewok 9 Solutions Execises fom Chapte 3: 3.3, 3.8, 3.15, 3.19, 3.0, 5.11, 5.1, 5.13 Execise 3.3. Suppose that C and C ae two cicles with the same cente q and adii,. Show that the composition i C C is the dilation D q,λ aound q by a facto of λ, whee λ = ( ). Solution. We have i C (p) = q + so that p q ) i C i C (p) = i C (q + ( ) = q + ( ) q + p q q ( ) = q + p q ( ) = q + ( ) p q 4 ( ) = q +. ) ) ((q + q This last expession is exactly the dilation of p aound q by the facto ( /), as desied. Execise 3.8. If x, y R and C is a cicle, show that thee is a unique licle l in R that passes though x, y and is othogonal to C. Solution. If x is a point on a licle S othogonal to C, then by Coollay 3.7 S is fixed by the invesion i C so that i C (x) must also be on S. Since thee is a unique licle between thee distinct points of R, thee is a unique licle though x, y, i C (x) povided that x is not on C. By Coollay 3.6(b) this licle is othogonal to C as desied. If both x, y ae on C, then the tangents to C at x, y eithe intesect at 1

2 a unique point q so that the cicle centeed at q with adius x q is the unique cicle othogonal to C and though x, y, o the tangents ae paallel, in which case x and y lie on a unique diamete of C that extends to a unique licle containing x, y and othogonal to C. Execise Suppose that C 1, C, C 3 ae all tangent, but not all at the same point. Show that thee ae exactly two Apollonian cicles D 1, D associated to C 1, C, C 3 at thee diffeent points. Solution. Choose one of the points of tangency, say the point q in C 1 C, and choose a cicle C centeed at q. Since the cicles C 1 and C pass though q, the invesion ι C takes them both to staight lines not though q. Since these cicles intesect only at q, thei images ι C (C 1 ) and ι C (C ) ae disjoint, i.e. paallel. Since C 3 doesn t pass though q, its image ι C (C 3 ) is a cicle not though q. Moeove, since ι C is confomal, the image ι C (C 3 ) is tangent to the paallel lines ι C (C 1 ) and ι C (C ). We claim that it suffices to show: Given thee licles that ae a pai of paallel lines and a cicle tangent to both lines, thee ae ae exactly two cicles tangent to these thee licles. In this case, the image unde ι C of these two cicles is a pai of licles that ae each tangent to C 1, C, C 3 simultaneously. Rotating and tanslating the plane if necessay (these isometies of the plane peseve lines and cicles), we can assume the two staight lines ae x = 1 and x = 1. Now any cicle simultaneously tangent to both lines must have cente on the line x = 0 and adius 1. Given one such cicle, it s clea that thee ae only two othe cicle tangent to this one and both lines. Execise Let C, D be non-intesecting cicles, and l be a line though thei centes. Show that thee is a cicle E centeed on l that is othogonal to C, D. Solution. Suppose that D is contained in C as pictued in the text on p. 105, and let p(x) denote the point p along l at distance x fom the cicle C. Let the adius of C be given by R and that of D by, and let c indicate the distance between the centes of C and D. Given the point p(x) on the line l, the length of the segment fom p(x) tangent

3 3 to the cicle C is given by f(x) = (x + R) R and the length of the segment fom p tangent to D is given by g(x) = (x + R c). Note that since C and D ae disjoint, we must have R c >. Thus (R c) > 0 and we have Moeove, we have f(0) = 0 < (R c) = g(0). f(x) g(x) = (x + R) R (x + R c) + = (x + R) R (x + R) + c(x + R) c + = c(x + ) + R c. Evidently, f(x) g(x) + as x +. In paticula, fo x lage enough, f(x) g(x) > 0. Since both f(x) and g(x) ae positive, this implies that f(x) > g(x). Conside the function f(x) g(x). When x = 0, this function is negative, while fo x lage enough this function is positive. By continuity thee is some value x 0 fo which f(x 0 ) = g(x 0 ). The cicle centeed at p(x 0 ) of adius f(x 0 ) (which is equal to g(x 0 )) is othogonal to both C and D, as desied. The case whee the cicles ae not nested is simila. Execise 3.0. Pove that if C, D ae non-intesecting cicles in R, thee is some cicle S R such that ι S (C) and ι S (D) ae concentic cicles. Solution. Suppose that C, D ae non-concentic cicles, and let l indicate the staight line joining the centes of C and D. Because l passes though the centes of C and D, it must be othogonal to both C and D. By Execise 3.19 thee is a cicle E, centeed on l, that is also othogonal to both C and D. Conside an intesection point p of E and l, and let S be a cicle centeed at p. Note that the cente of S lies on l, so S and l ae othogonal. We have now that both cicles C and D ae othogonal to the pai of othogonal licles S and l. Because E is a cicle that passes though the cente of S, the invesion ι S sends E to a staight line ι S (E); likewise because the line l is othogonal to S we must have ι S (l) = l. Thus the pai of images ι S (E) and ι S (l) ae a pai of othogonal lines. The cicles C and D don t pass though the point p, so thei images ae a pai of cicles ι S (C)

4 4 and ι S (D) that ae othogonal to both of the othogonal lines ι S (E) and ι S (l). This implies that ι S (C) and ι S (D) that ae centeed at the intesection of the lines ι S (E) and ι S (l), as desied. Execise You and a fiend walk upwads along the vetical half lines x = a and x = b at unit hypebolic speed. Paametically, you paths ae α(t) = (a, e t ), and β(t) = (b, e t ). Show that the distance between you and you fiend at time t satisfies d H (α(t), β(t)) b a e t. Solution. Suppose without loss of geneality that b > a. The hoizontal path γ y (t) = (t, y) fo t [a, b] has length Thus we have as desied. l H (γ y ) = b a γ y(t) (γ y ) (t) dt = b a dt y = b a y d H (α(t), β(t)) l H (γ e t) = b a e t, Execise 5.1. Suppose that l, l shae an endpoint on the x-axis. Show that thee is some hypebolic line m such that R m (l) and R m (l ) ae vetical half lines, and use this and the pevious execise to show that l and l ae asymptotic.. Solution. Let q indicate the shaed endpoint of l and l, and let m be a hypebolic line given by the intesection of a cicle C centeed at q with H. The hypebolic eflection R m is the estiction of the invesion ι C to H. Since l and l ae cicles though q, the cente of C, the invesions ι C (l) and ι C (l ) ae contained in staight lines (as opposed to cicles). Since the hypebolic eflection R m takes hypebolic lines to hypebolic lines, the images R m (l) and R m (l ) must be a pai of vetical lines. By Execise 5.11, thee ae unit-speed paametizations α and β of R m (l) and R m (l ) espectively so that the distance between α(t) and β(t) goes to zeo exponentially fast. Since R m is an isomety,

5 we have d H (R m α(t), R m β(t)) = d H (α(t), β(t)). Thus we have the unit-speed paametizations R m α and R m β of l and l espectively so that the distance between R m α and R m β goes to zeo exponentially fast. In paticula, l and l ae asymptotic. Execise Show that if p H and v is a vecto based at p, thee is a unique hypebolic line l passing though p tangent to v. 5 Solution. Note fist that if v is vetical, it is tangent to the vetical hypebolic line though p. Moeove, no othe hypebolic line though p can be tangent to v: The tangent line to a cicle centeed on the x-axis is vetical only at the pai of intesection points with the x-axis. Note that a vecto at p is tangent to a cicle if and only if the cicle passes though p, and the cente of C is on a line though p pependicula to v. If v is not vetical, then conside the line though p pependicula to v. Since v is not vetical, this line is not hoizontal, so it must intesect the x-axis, say at the point z. The cicle C of adius z p and cente z passes though p tangent to v, and so the intesection of C with H is the desied hypebolic line. Uniqueness follows as well, since any othe such hypebolic line must be contained in a cicle whose cente lies on the line though p pependicula to v, and its cente must also lie on the x-axis since it is a hypebolic line. Thus its cente is z and its adius is z p.