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(7) Find part of the locus of points at which two adjacent sides of a square subtend equal angles.
(8) Find the locus of a point at which two adjacent sides of a rectangle subtend supplementary angles.
INTERSECTION OF Loci.
When a point has to be found which satisfies two conditions, the problem is generally determinate if it is possible: and the method of loci is very frequently employed in discovering the point. For if the locus of points which satisfy each condition separately is constructed, it is obvious that the points which satisfy both conditions must be the points common to both loci, that is, must be the point or points where the loci intersect.
For example; a triangle is to be constructed on a given base with its sides of given lengths. Let AB be the base.
The two conditions are that the lengths of the two sides are given; the point sought for is the vertex: now the vertex must be at a certain distance from A= one of the given lengths; its locus is therefore a certain circle round A as centre. Similarly it must be at a certain distance from B; its locus is therefore another circle round B as centre. The points of intersection of these circles are therefore the vertices of the two equal triangles which fulfil the given conditions.
It was this reasoning that suggested the construction in Problem 5.
Occasionally it will be found that with certain conditions among the data in the following Exercises the loci do not intersect, or the solution becomes impossible. So in the case given, it will not be difficult to see that the circles would not intersect unless any two of the given sides were greater than the third side. These conditions among the data for the possibility or impossibility of a solution should always be found.
The principle of the intersection of loci may be thus stated :
If A is the locus of a point satisfying the condition X, and B the locus of a point satisfying the condition Y; then the intersections of A and B, and these points only, satisfy both the conditions X and Y.
The following examples of intersection of loci are important, and are at once demonstrated by the aid of the preceding examples of loci.
i. There is one and only one point in a plane which is equidistant from three given points not in the same straight line.
ii. There are four and only four points in a plane each of which is equidistant from three given straight lines that intersect one another but not in the same point.
EXERCISES ON INTERSECTION OF LOCI.
1. Find a point in a given straight line at equal distances from two given points. Construct the figures for all cases.
2. Find a point in a given straight line at a given distance from a given straight line.
3. Find a point in a given straight line at equal distances from two other straight lines.
4. On a given straight line to describe an equilateral triangle.
5. Describe an isosceles triangle on a given base, each of whose sides shall be double of the base.
6. Find a point at a given distance from a given point, and at the same distance from a given straight line.
7. Given base, sum of sides, and one of the angles at the base, construct the triangle.
8. Given base, difference of sides, and one of the angles at the base, construct the triangle.
9. Find a point at a given distance from the circumference of two given circles, the distances being measured along their radii or their radii produced.
10. A straight railway passes within a mile of a town. A place is described as four miles from the town, and half a mile from the railway. How many places satisfy the conditions ?
11. Find a point equidistant from three given straight lines that intersect one another but not in the same point.
ANALYSIS AND SYNTHESIS.
If problems cannot be solved by this method, it remains to attack them by the method, as it is called, of Analysis and Synthesis. This is not so much a method as a way of searching for a suggestion, and nothing but experience and ingenuity will here avail the student. The solution is
supposed to be effected, and relations among the parts of the figure are then traced until some relation is discovered which can give a clue to the construction. Nothing but seeing examples can make this clear.
(1) It is required to draw a line to pass through a given point and make equal angles with two given intersecting lines.
Let O be the given point, AB, AC the given lines.
We reason as follows (analysis): suppose POQ were the line required, then the angle at P= angle at Q.
Therefore AP=AQ; therefore if we bisected the angle A, POQ would be at right angles to the bisector.
Now this is a suggestion we can work backwards from, and the construction is as follows.
Synthesis. Bisect the angle BAC, and let fall OH a perpendicular to the bisector, and let it meet the lines in P, Q, and POQ can then be proved to be the line required.
(2) It is required to draw from a given point three straight lines of given lengths, so that their extremities may be in the same straight line, and intercept equal distances on that line.
Analysis. Suppose OA, OB, OC were the three lines, so that CBA is a straight line, and CB = BA. . Then it occurs. to us that if so
OB were prolonged to D, making BD=OB, then CD and DA would be respectively parallel and equal to OA and OC (see $ 3, Ex. 21); and that the sides of the triangle DOA are respectively equal to OA, OC and 20B. Hence the construction is suggested.
Synthesis. Make a triangle DOA whose sides are OA, OC, and 20B; complete the figure, by drawing DC, OC parallel tò OA, AD; and the other diagonal ABC will be the line required. For it may be shewn that AB = BC.
The student must not be surprised if he finds problems of this class difficult. For there is nothing except previous knowledge of geometrical facts to point out which of the many relations of the parts of the figure are to be followed up in order to arrive at the particular relation which suggests the construction. It is not easy to see what is to suggest the producing of OB to D as in the figure.
Subjoined are a few problems of no great difficulty, which may be solved by this method.
1. On a given straight line to describe a square.
3. Given two sides of a parallelogram and the included angle, construct the parallelogram.
4. Given the lengths of the two diagonals of a rhombus, construct it.
5. From a given point without a given straight line to draw a line making an angle with the line equal to a given angle.