Calcolo geometrico, secondo l'Ausdehnungslehre di H. by Peano G.

By Peano G.

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M Since the sums + m, m + n, n + are all positive, the argument remains valid, as long as we apply the correct signs to the various segments and areas involved. (6) = n+ 59 2◦ . If the sums + m, m + n, n + are all negative, then the argument above shows that the expression s takes on a maximal value at the point O we have constructed. Indeed, equation (6) shows that, for example, if we fix K , L , a smaller length for J J1 results in a larger value for s. 3◦ . In this case, s has neither a maximum nor a minimum, for any triangle J K L .

For 43 the same reason, N L retains its direction. And since KO, LO are perpendicular to these two segments, these two sides of KN LO also retain their direction. We now show that N, K, L slide along fixed lines. Point N certainly slides along line BC. Point K also slides along a fixed line. Indeed, if M N and M N (not shown) are two corresponding sides of possible minimal quadrilaterals, then M N M N , and triangles M N B, M N B are homothetic. This the midpoints K, K of these segments lie along the median of any one of these triangles, so K slides along this median.

2o . Show that the bisector of angle HM K is parallel to a side of the rectangle. 3o . Find point M , knowing points H and K. 4o . This last problem has two solutions M, M . Show that the circle with diameter M M is orthogonal to the circumscribed circle of the rectangle. 5o . Find the locus of points M such that P R is perpendicular to QS. Figure t360a Solution. ) We will show that these fixed lines are the diagonals of the given rectangle. Suppose the rectangle is ABCD (fig. t360a), O the intersection of its diagonals, and H the intersection of lines BD, P R.

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