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Chapter 3 110

# Chapter 3 110 - 260 25 26 27 3 CHAPTER3 PROBLEMS PLUS y =...

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Unformatted text preview: 260 25. 26. 27. 3 CHAPTER3 PROBLEMS PLUS y = x4 — 2:132 — a: => 3/ = 4x3 — 4m — 1. The equation of the tangent line at a: = a is y — (a4 — 2(12 — a) — (4a3 4a 1)(ar a) ory — (4(13 — 4a —1)av + (—3114 + 2a2) and similarly for as = b. So if at m : a and m : b we have the same tangent line. then 4a3 # 4a # 1 : 4b3 — 4b — 1 and —3a4 + 2a2 2 —3b4 + 2b2. The ﬁrst equation gives a3 — b3 = a — b => ((1 — b)(a2 + ab + b2) : (a — b). Assuming a 3A b. we have 1 2 a2 + ab + b2. The second equation gives 3(a4 # b4) : 2(a2 — b2) => 3(a2 — b2)(a2 + b2) : 2(a2 — b2) which is true if a z —b. Substituting into 1 : a2 + ab + b2 gives 1: a2 A a2 + a2 => a : :l:1 so that a = 1 and b = —1 or vice versa. Thus. the points (17 —2) and (—10) have a common tangent line. As long as there are only two such points. we are done. So we show that these are in fact the only two such points. Suppose that a2 — b2 7é 0‘ Then 3(a2 — b2)(a2 + b2) : 2(a2 — b2) gives 3(a2 + b2) : 2 or a2 + b2 : %. 1 Thus,ab: (a2+ab+b2)—(a2+b2) :1—%: %,sob: \$.Hence.a2+§% : §.so9a4—t—1:6a2 => 1 0 : 9a4 — 6a2 + 1 : (3a2 7 1)2. So 3(12 1 0 a2 g s b2 9112 % a2, contradicting our assumption that a2 75 b2. Suppose that the normal lines at the three points (a1, (1%). (a2, a2). and (a3, a3) intersect at a common point. Now if one of the ai is 0 (suppose a1 : 0) then by symmetry (12 : —a3, so all + a2 + a3 = 0. So we can assume that none of the ai is 0. The slope of the tangent llne at (0411,01?) IS 2m. so the slope of the normal line 1S ’2— and its equation is (12‘ 1 . , . . y — a? = #_2_ (a: — a7). We solve for the m—coordmate of the 1ntersection of the normal lines from ((11, a?) (1i 1 1 d . 2 : = 2 — -— — = ,2 7 — f :> an (@2112) y a1 2&1 (:0 a1) a2 2112 (er a2) 1 1 2 2 a1 — (12 x —- 7 — : a2 — (11 => :0 2 (—a1 — a2)(a1 + a2) <1} 3: : —2a1a2(a1 + a2) (7%). 2(12 2a1 2a1a2 Similarly, solving for the m-coordinate of the intersections of the normal lines from (a1, a?) and (L137 a3) gives a: : #2a1a3(ai + (13) (T). Equating (7k) and (T) gives a2(a1 + a2) = a3(a1 + a3) 4:» a1(a2 — a3) : (13, — a3 : —(a2 + a3)(a2 — a3) a) a1:#(a2+a3) 41> ai+a2+a3:0. Because of the periodic nature of the lattice points. it sufﬁces to consider the points in the 5 X 2 grid shown. We can see that the minimum value of r occurs when there is a line with slope % which touches the circle centered at (3, 1) ...
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