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4.17 The ancient Greeks classified numbers geometrically. For example, a number was called trian-

Fig. 3-3

gular if that number of pebbles could be arranged in a symmetric triangle. The first eight triangular numbers are 1, 3,6, 10, 15,21,28, and 36:

PRINCIPLE 3:

T,= 1

(Angle-Side-Angle, ASA) If two angles and the included side of one triangle are congruent to the corresponding parts of another, then the triangles are congruent.

T, = 3

Fig. 3-4

T, = 6

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This structure will result in 10 passes through the loop. Note that count is assigned a value of 1 before entering the loop. The value of count is then incremented by 1 during each pass through the loop. Once the value of count exceeds 10, the execution will cease.

PRINCIPLE 4:

T,= 10

T,= 15

(Side-Side-Side, SSS) If three sides of one triangle are congruent to three sides of another, then the triangles are congruent.

Write and test the boolean function:

Fig. 3-5

int isTriangular(int n)

Selecting congruent triangles From each set of three triangles in Fig. 3-6 select the congruent triangles and state the congruency principle that is involved.

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The argument n is triangular if and only if it is a sum of consecutive integers 1 + 2 + 3 + . So we just have to compute these sums until we find one that is greater than or equal to n. If that sum is equal to n, then n is a triangular number; otherwise, it isn t:

(a) nI (b) nII (c) nI nII, by SAS. In nIII, the right angle is not between 3 and 4. nIII, by ASA. In nI, side 10 is not between 70 and 30 . nII nIII by SSS.

int isTriangular(int); main0 int n; do -t tin >> n; if (isTriangular( tout -CC n << ' is triangular.\n"; else tout CC n << ' is not triangular.\n"; } while (n > 0);

CHAP. 3]

// Returns 1 i f n i s a triangular int isTriangular(int n) 1 int i = 0, sum = 0; while (sum < n) sum += ++i ; if (sum == n) return 1; else return 0;

Fig. 3-6

number (1,

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Determining the reason for congruency of triangles In each part of Fig. 3-7, nI can be proved congruent of nII. Make a diagram showing the equal parts of both triangles and state the congruency principle that is involved.

10, 15, etc.):

Fig. 3-7

We assume that the max ( int , int max(int, int); in t ) function is already available:

Fig. 3-8

int max(int x, int y, int z) -t int max(int,int); return max(max(x,y),z); )

Finding parts needed to prove triangles congruent State the additional parts needed to prove nI nII in the given figure by the given congruency principle.

Every point in the coordinate plane has a unique pair (x, y) of rectangular coordinates and a unique pair (r, 0) of polar coordinates with r 2 0 and 0 5 0 < 27~. The following function converts from rectangular to polar coordinates. Since the output consists of more than one variable, the two output variables r and t are passed by reference:

If we choose to use an Until clause rather than a While clause, we can write the control structure in either of the following ways.

void rectangularToPolar(double&

Fig. 3-9

r, double& t, double x, double y)

(a) In Fig. 3-9(a) by SSS. (b) In Fig. 3-9(a) by SAS. (c) In Fig. 3-9(b) by ASA. (d) In Fig. 3-9(c) by ASA. (e) In Fig. 3-9(c) by SAS.

const double pi = 3.1415926535897932385; r = sqrt(x*x + y*y); if (x > 0) if (y >= 0) t = atan(y/x); else t = atan(y/x) + 2*pi; else if (x == 0) if (y > 0) t = pi/2; else if (y == 0) t = 0;. else t = 3*pi/2; else t = atan(y/x) + pi;

(a) If AD > BC, then nI (b) If /1 /4, then nI nII by SSS. nII by SAS. nII by ASA. nII by ASA. nII by SAS.

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