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Ex. 1002. Given lines AB and p. In the line AB, to find a point C so that ACP CB? = p2.

Ex. 1003. In the line AB, to find a point C so that ACP = 2 CB2.

Ex. 1004. In the median AD of A ABC, to find a point X such that XD and the perpendiculars dropped from X upon AB and AC divide the figure into three equivalent parts.

Ex. 1005. In a given square ABCD, to inscribe an equilateral triangle, having one vertex at A.

Ex. 1006. From a point P without a circumference, to draw a secant which is bisected by the circumference.

Ex. 1007. In a given square, to inscribe another square, having a given side.

Ex. 1008. To inscribe a square in a semicircle.
Ex. 1009. In the triangle ABC, to in-

B
scribe a parallelogram having a given area,
and having an angle common with A 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

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line p.

D

Ex. 1011. To construct a triangle, having given the base, the vertical angle, and the bisector of that angle. (Ex. 1010.)

Ex. 1012. Upon a given line as hypotenuse to construct a right triangle one of whose arms is a mean proportional between the other arm and the hypotenuse.

Ex. 1013. To transform a given square into a rectangle having a perimeter equal to twice the perimeter of the given square.

Ex. 1014. Given two concentric circles. To draw a chord in the larger 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.

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Hyp. In A ABC and DFE, AB=DF, AC=DE, ZA=rt. Z, and < D is oblique.

area ABC > area DEF.

To prove

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..A ABC and DEF have equal bases and unequal altitudes.

A ABC > A DEF.

(Why?)

.. A ABC is a maximum.

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.

В.

А
Нур. . A ADC = A ABC, and AB = BC.
To prove AB + BC + AC < AD + DC + AC.
Proof. Produce AB by its own length to E. Join BD
and DE.

BD | AC,
(for otherwise A ABC would not be A ADC).
ZEBD=ZBAC.

(Why?) < BAC = _ BCA.

(Why?) Z BCA=CBD.

(Why?) .. ZEBD=2 CBD.

(Ax. 1.) But

BE,

(Why ?) and

BD is common.
.. ABDC=ABDE.

.:. DC= DE.
But
AD + DE > AB + BE.

(Why ?) . AD + DC > AB + BC, or adding AC,

AC + AD + DC > AB + BC + AC.

BC=

Q.E.D.

PROPOSITION III. THEOREM 443. The maximum of isoperimetric triangles on the same base is the one whose other two sides are equal.

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To prove

Hyp. A ABC and ABD have equal perimeters, and AC=CB.

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?)

.. AF= FB. But

A AFB = A ADB,

(same base, equal altitudes.) .. perimeter AFB < perimeter ADB. (442)

.. AF+ FB < AD + DB or AC + CB. Hence

AF < AC. .. FE < CE.

(311) .. area AFB < area ACB,

(unequal altitudes),
area ADB < area ACB.

or

Q.E.D.

444. Cor. Of all isoperimetric triangles, the equilateral has the maximum area.

PROPOSITION IV. THEOREM

445. Of all polygons having all sides given but one, the maximum can be inscribed in a semicircle having the undetermined side as diameter.

E

F

C

B

Hyp. Polygon ABCDEF is the maximum of all polygons having given the sides AF, FE, ED, DC, and CB.

To prove ABCDEF can be inscribed in a semicircle whose diameter is AB.

Proof. Join any vertex, as D, with A and B.

Then A ADB must be the maximum of all triangles that can be formed with sides AD and DB. For otherwise by making ZADB a right one without changing the sides AD and DB, we could increase A ADB without altering the remaining parts AFED and DCB of the polygon. Or polygon ABCDEF would be increased, which is contrary to the hypothesis, since ABCDEF is a maximum.

.: A ADB is the maximum of all triangles having AD and DB given. Then LADB 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.

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