EXERCISES OF GREATER DIFFICULTY

Ex. 934. Construct a rectangle equivalent to a given square and having the sum of its base and altitude equal to a given line.

Ex. 935. Construct a rectangle equivalent to a given triangle and having its perimeter equal to a given line.

Ex. 936. Construct two lines, having given their sum and their product.

Ex. 937. Construct a rectangle equivalent to a given square and having the difference of its base and altitude equal to a given line.

Ex. 938. Construct a rectangle equivalent to a given rhombus, the difference of the base and altitude of the rectangle being equal to a given line.

Ex. 939. Construct two lines, having given their difference and their product.

Con

Ex. 940. Construct a triangle, given its three altitudes. HINT. struct first a triangle whose sides are proportional to the three given altitudes.

Ex. 941. Through the vertices of an equilateral triangle draw three lines which shall form an equilateral triangle whose side is equal to a given line,

Ex. 942. The feet of the perpendiculars dropped upon the sides of a triangle from any point in the circumference of the circumscribed circle are collinear.

OUTLINE OF PROOF. The circle having AP as diameter will pass through M and Q. ... 21 = 21' and 2=2'. Similarly

K.

B

... ZAPQ = Z BPK.

.. ZAMQ = Z BMK.

... QM and MK form one str. line.

Q

C

Ex. 943. Given the base, the angle at the vertex, and the sum of the other two sides of a triangle; construct the triangle.

ANALYSIS. Imagine the problem solved and draw ▲ ABC. Prolong AB, making BE = BC, since the line ca is given. Since ZE=}ZABC, ▲ AEC can be constructed.

Ex. 944. The hypotenuse of a right triangle is given in magnitude and position; find the locus of the center of the inscribed circle.

B

E

A

Ex. 945. Prove Prop. VIII, Book IV, by using the following figure, in which A'D'E' is placed in the position ADE.

Ex. 946. Prove Prop. VIII, Book IV, using two triangles such that one will not fall wholly within the other.

Ex. 947. If two triangles have an angle of one supplementary to an angle of the other, the triangles are to each other as the products of the sides including the supplementary angles. (Prove by method similar to that of Ex. 945.)

Ex. 948. Given base, difference of sides, and difference of base angles; construct the triangle.

ANALYSIS. In the accompanying figure suppose c and EA, (b− a), to be given. A consideration of the figure will show that ≤2=1+ZA. Add 1 to both members of the equation; then 21 +22 221 + ZA. But 21 22

=

LB = 221 + ZA.

LB.

B

... 21 = }} (ZB – ZA).

C

E

A

The A BEA may now be constructed. The rest of the construction is left for the student.

Ex. 949. Given base, vertex angle, and difference of sides, construct the triangle.

Ex. 950. If upon the sides of any triangle equilateral triangles are constructed, the lines joining the centers of the equilateral triangles form an equilateral triangle.

E

B

HINT. Circumscribe circles about the three A equilateral A. Join O, the common point of

C

intersection of the three circles, to A, B, and C, the vertices of the given §. Prove each at O the supplement of the opposite in an equilateral ▲, and also the supplement of the opposite in the A to be proved equilateral.

Ex. 951. To inscribe a square in a semicircle.

ANALYSIS. Imagine the problem solved and ABCD the required square. Prove OA = AB. Draw EF1 EO

2

meeting OB prolonged at F. FE: EO BA: AO. ... EF 2 OE.

=

Ex. 952. To inscribe a square in a given triangle.

R

B

E A OD

A KD TH

OUTLINE OF SOLUTION. Imagine the problem solved and ABCD the required square. Draw SF RT and construct square KSFH. Draw RF, thus determining point C. The cons. will be evident from the figure. To prove ABCD a square, prove BC CD. BC: SF CD: FH. But SF FH. =

Ex. 953. In a given square construct a square, having a given side, so that its vertices shall lie in the sides of the given square.

HINT. Ex. 954. Construct a triangle, given ma, mc, hb. ANALYSIS. Imagine the problem solved and that ABC is the required A. If BF were moved I to itself till it contained K, the rt. ▲ AKT would be formed and KT would equal ¦ BF.

Construct a rt. A, given the hypotenuse and sum of arms.

B

E

K

Then AAKT can be made the basis of the required construction.

A

B

8

K

d

8'

Ex. 955. Construct a quadrilateral, given two of its opposite sides, its two diagonals, and the angle between them. OUTLINE OF CONSTRUCTION. Imagine the problem solved and that ABCD is the required quadrilateral, s, s', d, d', and ▲ DOC being given. By motion of d and d' the parallelogram BKFD may be obtained. The required construction may be begun by drawing BKFD, since two sides and the inIcluded

are known. With K as center and s

D

as radius, describe an arc; with D as center and s' as radius, describe an arc intersecting the first, as at C. Construct ABKC, or □ DACF, to locate A.

Ex. 956.

Between two circles draw a line which shall be parallel to

Ex. 958. Transform any given triangle into an equilateral triangle by a method different from that used in Ex. 865.

Call the base of the given triangle b and its altitude a. X = the side of the required equilateral triangle.

BOOK V

REGULAR POLYGONS. MEASUREMENT OF THE

CIRCLE

515. Def. A regular polygon is one which is both equilateral and equiangular.

Ex. 959. Draw an equilateral triangle. Is it a regular polygon?

Ex. 960. Draw a quadrilateral that is equilateral but not equiangular; equiangular but not equilateral; neither equilateral nor equiangular; both equilateral and equiangular. Which of these quadrilaterals is a regular polygon?

Ex. 961. Find the number of degrees in an angle of a regular dodecagon.

516. Historical Note. The following theorem presupposes the possibility of dividing the circumference into a number of equal arcs. The actual division cannot be obtained by the methods of elementary geometry, except in certain special cases which will be discussed later.

As early as Euclid's time it was known that the angular magnitude about a point (and hence a circumference) could be divided into 2", 2n. 3, 2′′. 5, 2n. 15 equal angles. In 1796 it was discovered by Gauss, then nineteen years of age, that a regular polygon of 17 sides can be constructed by means of ruler and compasses, and that in general it is possible to construct all polygons having (2" + 1) sides, n being an integer and (2" + 1) a prime number. The first four numbers satisfying this condition are 3, 5, 17, 257. Gauss proved also that polygons having a number of sides equal to the product of two or more different numbers of this series can be constructed.

Gauss proved, moreover, that only a limited class of regular polygons are constructible by elementary geometry. For a note on the life of Gauss, see § 520.