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LOCUS PROBLEMS

138. PROBLEM 17. To find the locus of a point such that the perpendiculars from it to two given lines shall have a given ratio.

Let OR and OS be the given lines. Let P be a point such that the perpendiculars PM and PN have a given ratio ▲ : B.

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To find the locus of the point P.

As in I. 254 the point o is a point on the locus. Another point of the locus is obtained by finding Q such that the perpendiculars QR, QS are respectively equal to A, B (I. 257, ex. 2).

Now the line QP must pass through 0, because the broken lines RQS and MPN are similar and similarly placed (72, ex. 1). Hence any point P, situated in the angle ROS and satisfying the given condition, lies on the fixed line oQ. Conversely, any point on the line oQ satisfies the given condition. [The proof is left to the student.]

Show that there is another part of the locus.

Ex. Find a point from which the perpendiculars to three given lines shall have given ratios L: M: N. How many solutions are there?

139. PROBLEM 18. To find the locus of a point such that its joins to two given points shall have a given ratio.

Let A and B be the given points, and H: K the given ratio.

Let P be a point such

that PA: PB = H: K.

To find the locus of P.

The points M and N which divide AB internally and externally in the ratio H: K are evidently points on the locus.

The line PM bisects

N

A M

B

the angle APB, because PA: PB =AM: MB.

[54

Similarly PN bisects the external angle between PA and PB. Then the angle MPN is a right angle (I. 100, ex. 1). Hence P is on the circle whose diameter is MN (III. 57). It follows that any point satisfying the given condition lies on this circle.

Conversely, to prove that every point on this circle satisfies the given condition.

Let P be any point on the circle.

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Draw PA' making the angle MPA' equal to MPB, and meeting AB in some point a', which is to be proved coincident with 4.

Since MPN is a right angle, hence PN bisects the external angle between PA' and PB. Therefore A' is the harmonic conjugate of B with regard to M and N (59).

But 4 is also the harmonic conjugate of B with regard to M and N. Therefore A' coincides with 4 (61).

Hence PM bisects the angle APB; and therefore

PA: PB AM : MB H: K

Ex. 1. Compare the position of the locus in the three cases

H<=>K.

Ex. 2. Find the position of a point, whose joins to three given points have given ratios H: K:L; show that there may be two solutions, one solution, or none.

EXERCISES

1. The rectangle of two lines is a mean proportional between the squares on the lines.

2. Find a line such that the perpendiculars to it from three given points may have given ratios to each other.

3. A regular polygon inscribed in a circle is a mean proportional between the inscribed and circumscribed circles of half the number of sides.

4. Construct a triangle, being given its base, ratio of sides, and either altitude or vertical angle.

5. Find the locus of a point at which two given circles shall subtend equal angles.

6. Two diagonals of a regular pentagon divide each other in extreme and mean ratio.

Definition. Two points P and P' are said to be similarly situated (or to correspond) with regard to two circles, whose centers are Cand C', when CP and C'P' are parallel and in the ratio of the radii. The correspondence is called direct or transverse according as OP and O'P' are at the same or opposite sides of the central line. The following exercises illustrate the theory of correspondence.

7. The line PP' divides the central line either externally or internally in the ratio of the radii, and each of the points of division is a self-corresponding point (called a center of similitude).

8. Two polygons whose respective vertices are all in direct (or transverse) correspondence are similar and have the point S (or S') for center of similitude.

9. The line joining a fixed point to a variable point on a fixed circle is divided in a constant ratio; prove that the locus of the point of division is a circle, and that the two circles have the fixed point as center of similitude.

10. In a given sector OAB inscribe a square, so that two corners may be on the arc AB. [Take any square having a side parallel to AB; circumscribe it by a sector having its radii parallel to OA and OB; then use the principle of correspondence.]

11. A common tangent passes through a center of similitude.

12. Describe a circle through a given point (P) to touch two given lines (OA, OB). [Draw any circle touching the two lines; let it meet OP in P'; then considering P and P' as corresponding points, find the center of the required circle. Two solutions.]

BOOK VI. -MENSURATION

1. Mensuration is the science of measurement. The operation of measuring a magnitude by means of another magnitude of the same kind will be defined after certain preliminary notions are explained. It will be shown to be intimately connected with the theory of ratio set forth in Book IV and applied in Book V.

In Art. 11 of Book IV the scale of relation of two magnitudes of the same kind was explained, and it was shown that any two such magnitudes have a definite ascending order in which their various multiples occur. The theorems of Books IV and V are based on the mere fact that this scale is definite, and their proofs do not require the actual determination of any particular scale of relation.

The determination of a scale (or of a selected portion of it) is, however, important for other purposes, and is the fundamental problem in Mensuration.

ABBREVIATED SCALE

2. If two magnitudes A and B are commensurable, then some of their multiples are equivalent and occupy the same place in the ascending scale of magnitude. If any two multiples mA and B are known to be equivalent, then the ratio : B is completely defined, for it is equal to the ratio of two whole numbers n: m (IV. 43), and hence the order of any assigned multiples could be written down.

Any ratio that is equal to the ratio of two whole numbers is called a rational ratio; thus the ratio of two commensurable magnitudes is a rational ratio, and the ratio of two incommensurable magnitudes is an irrational ratio.

If two magnitudes A and B are incommensurable, it will be proved presently that their ratio can be sufficiently characterized by assigning the intervals in which merely the decimal multiples of the antecedent (4, 10 4, 100 A, ...) are found among all the multiples of the consequent (B, 2B, 3 B,...). Such an arrangement is called the abbreviated scale of A and B.

215 B, 100 A, 216 B,

The following is an example of an abbreviated scale:
2 B, A, 3B,... 21 B, 10 A, 22 B,
2159 B, 1000 4, 2160 B,

....

...

This exhibits the position of the decimal multiples of the antecedent, showing that

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1000 A lies between 2159 B and 2160 B.

The abbreviated scale of P and Q is said to be similar to that of A and B if all the like decimal multiples of P and A lie between like multiples of Q and B.

E.g., if the abbreviated scale of P and Q is

2Q, P, 3 Q, ..., 21 Q, 10 P, 22 Q, ...,

then the abbreviated scale of P and Q is said to be similar to that of A and B written above.

Similar scales.

3. THEOREM 1. If the abbreviated scale of A and B is similar to the abbreviated scale of P and Q, then the complete scales are similar, that is to say, the ratios A: B and P: Q are equal.

For suppose, if possible, that the complete scales are somewhere unlike, then, by definition, the two ratios are unequal, say

A: BP: Q.

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