Ex. 1002. Given lines AB and p. In the line AB, to find a po so that AC2 - CB2 = p2. Ex. 1003. In the line AB, to find a point C so that AC2 = 2 CB Ex. 1004. In the median AD of ▲ ABC, to find a point X such XD and the perpendiculars dropped from X upon AB and AC d the figure into three equivalent parts. Ex. 1005. In a given square ABCD, to inscribe an equilateral tria having one vertex at A. Ex. 1006. From a point P without a circumference, to draw a se which is bisected by the circumference. Ex. 1007. In a given square, to inscribe another square, having a g side. Ex. 1008. To inscribe a square in a semicircle. Ex. 1009. In the triangle ABC, to inscribe a parallelogram having a given area, and having an angle common with ▲ ABC. Ex. 1010. From the midpoint A of arc BC, to draw a chord AD, intersecting chord BC in E, so that DE is equal to a given A line p. Ex. 1011. To construct a triangle, having given the base, the vertical angle, and the bisector of that angle. (Ex. 1010.) B E Ex. 1012. Upon a given line as hypotenuse to construct a right triang one of whose arms is a mean proportional between the other arm and t hypotenuse. Ex. 1013. To transform a given square into a rectangle having a p rimeter equal to twice the perimeter of the given square. Ex. 1014. Given two concentric circles. To draw a chord in the larg circle so that it equals twice the chord formed in the smaller circle. MAXIMA AND MINIMA OF PLANE FIGURES 439. DEF. A maximum is the greatest of all magnitudes of the same kind; a minimum, the smallest. 440. DEF. Isoperimetric figures are those which have equal perimeters. PROPOSITION I. THEOREM 441. Of all triangles having given two sides, that in which those two sides are perpendicular to each other is a maximum. с F A B D Hyp. In ▲ ABC and DFE, AB=DF, AC=DE, ZA=rt. ≤, D is oblique. and .. ▲ ABC and DEF have equal bases and unequal altitudes. .. ΔΑΒΟ > ΔΙΕF. .. ▲ ABC is a maximum. (Why?) Q.E.D. Ex. 1015. To divide a given line into two parts so that the rectangle contained by the segments is a maximum. Ex. 1016. In the hypotenuse of a right triangle to find a point so that the sum of the squares of perpendiculars drawn from the point upon the arms is a minimum. PROPOSITION II. THEOREM 442. Of all triangles having the same base and equal areas, the isosceles triangle has the minimum perimeter. E B Hyp. ▲ ADC ≈ ▲ ABC, and AB = BC. To prove AB + BC + AC < AD + DC + AC. Proof. Produce AB by its own length to E. and DE. But and BD\\ AC, (for otherwise ▲ ABC would not be ≈▲ ADC). Join BD (Why?) (Why?) (Why?) (Ax. 1.) (Why?) (Why?) But or adding AC, AC+ AD + DC > AB + BC + AC. Q.E.D. 443. The maximum of isoperimetric triangles on the same base is the one whose other two sides are equal. Hyp. AABC and ABD have equal perimeters, and AC=CB. To prove area ACB> area ADB. Proof. Draw median CE and DF AB, meeting CE in F. Join FA and FB. Then CE is the perpendicular bisector of AB. (Why?) (442) or .. FECE. .. area AFB < area ACB, (unequal altitudes), area ADB < area ACB. (311) Q.E.D. PROPOSITION IV. THEOREM 445. Of all polygons having all sides given but the maximum can be inscribed in a semicircle ha the undetermined side as diameter. Hyp. Polygon ABCDEF is the maximum of all polyg having given the sides AF, FE, ED, DC, and CB. To prove ABCDEF can be inscribed in a semicircle wh diameter is AB. Proof. Join any vertex, as D, with A and B. Then ▲ ADB must be the maximum of all triangles that c be formed with sides AD and DB. For otherwise by maki ZADB a right one without changing the sides AD and D we could increase ▲ ADB without altering the remaining par AFED and DCB of the polygon. Or polygon ABCDE would be increased, which is contrary to the hypothesis, sin ABCDEF is a maximum. ..▲ ADB is the maximum of all triangles having AD an DB given. Then ZADB is a right angle, (Why? and D is on a semicircumference that can be constructed on AB For the same reason every vertex of the polygon must lie on the semicircumference. Q.E.D Ex. 1017. To inscribe an angle in a given semicircle so that the sum of its arms is a maximum. |