By Hatcher A.

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**Additional info for Algebraic topology. Errata (web draft, Nov. 2004)**

**Sample text**

37 Figure t362b Now point A is on line AB, so AD1 = AD. And A is also on line AC, so AD = AD2 . Because the angle at A in isosceles triangle AD1 D2 does not vary, the length of D1 D2 depends only on the length of AD1 = AD = AD2 , and is smallest when this length is shortest, which is when D is the foot of the altitude from A to BC. The problem is solved, but we can characterize the required triangle of minimal perimeter more precisely: it is the triangle formed by the feet of the three altitudes of ABC.

Alternate proof of (2◦ ): We can write equation (7) in the form: (bz − cy)2 + (cx − az)2 + (ay − bx)2 4S 2 + . a2 + b2 + c2 a2 + b2 + c2 Now the expression on the right is clearly minimal when each of the expressions in parentheses is zero; that is, when x : y : z = a : b : c. Translating this to the geometric situation, it means that the sum of the squares of the distances from a point to the sides of a triangle is minimal when these distances are proportional to the sides of the triangle. As argued earlier, this means that this point must coincide with point O of exercise 365, and the minimal triangle with triangle P QR.

O1 ω12 + O2 ω22 = Oω 2 . The result of 189 tells us that O1 B 2 = O1 R · O1 R = (O1 ω1 + ω1 R) · (O1 ω1 + ω1 R ) = (O1 ω1 + ω1 R) · (O1 ω1 − ω1 R) = O1 ω12 − ω1 R2 . In the same way, we have O2 D2 = O2 P · O2 P = O2 ω22 − ω2 P 2 . Adding, we find O1 B 2 + O2 D2 = O1 ω12 − ω1 R2 + O2 ω22 − ω2 P 2 . Using notes 3-5, we can derive from this the relation OC 2 + ωM 2 = Oω 2 , in which OC and ωM are the radii of the circles in question, while Oω is the distance between their centers. This last relation shows that the two circles are orthogonal.