By Marina Cohen
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Similarly, since m is the perpendicular bisector of the side AY, C is equidistant from the points A and Y so that BX = AB and CY = AC. MX = MA and NA = NY, Also, and by the SAS congruency theorem, we have A BMX = ABM A and From the Exterior Angle Theorem, we have A CNY = ACNA. 52 MISCELLANEOUS TOPICS and and also AB + BC + CA = XB 4- BC + CY = XY. , and perimeter AY. 17. Construct a triangle given the foot F of an altitude, the circumcenter 5, and the center N of the 9-point circle. Solution. Analysis Figure.
In the following diagram, the segment AB is of length 3. Construct all points on the line AB whose power with respect to to is 4. Solution. If a point P is on the line outside the circle then PA • PB = PA - (PA + 3) = 4. 25 Solving this quadratic for PA, we get PA = 1 or - 4. Hence, there are two points on the line AB and they can be constructed as follows: (1) Trisect the segment AB in the usual way to obtain a segment BC of length 1. (2) With center A and radius BC, draw an arc cutting the line AB at a point P outside the segment AB.
Since AE = 2EF, the areas are as shown in the figure below. B F Since 3BF = 2FC, we have 2 3 BF FC [ABF] [AFC] 3x 3+ y 38 AREA so that 9x = 6 + 2 y. Similarly, BF FC [BDF] [DFC] 1+x y so that 2y = 3 3x, and therefore x = § and [DFC] = y = 154 ' (b) We have 3x + 3 + AC [ABC] AD [ABD] 2x + 2 y _ 9 ~~ 4* 17. Given a rectangle, construct a square having the same area. and Solution. Given a rectangle with sides a and 6, construct the segments BC with AB = a and BC — b. Next, construct a semicircle with diameter AC and erect a perpendicular to AC at B, hitting the semicircle at D.