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Here is an equivalent question: Two circles of the same radius intersect at 60 degree angle(s). In what proportion do they divide each other?
[In fact there is some ambiguity in this problem, for - which of the angles between two tangents is 60 degrees?]

Problem. Two circles of the same radius intersect at the angle 2d/3. Express in degrees the smaller of the arcs contained between the intersection points.

Can I get some clarification please? There was a remark appended to the problem which clarified the problem a bit, but I'm still not completely certain about what is required to be found, nor what is given.

Thanks

Problem: Find the geometric locus of points from which the tangents drawn to a given circle are congruent to a given segment.

Do I understand this correctly? Let the given segment be 'a'. If I choose a point P in the plane, and draw two tangents to some circle, and mark the tangency points A and B, then if PA = PB = 'a', the point P satisfies the condition for membership of the locus of points?

If that's the case, then the problem is quite easy once one learns the solution for exercise 249.

Writing this helped me understand the nature of the problem.

I've completed all of the problems in this chapter except for the very first. Once I have some time (probably 2-3 weeks from now), I'll dedicate some time to posting solutions to exercises in chapters 2.1 and 2.2. So far, I've had a very great time learning elementary geometry with this book. If a solution set to all (or nearly all) problems existed, then I think that this would be more readily adopted by more educators, and therefore more students would receive a rich elementary geometry experience.

OK! So, the good news is that there are two fruitful approaches now: one based on comparing angles, the other based on the triangle inequality. Thanks!

1. AM < AP follows the same argument. Introducing N was unnecessary, yes.

2. I see it. The solution for AM < AP using the triangle inequality involved drawing 1 less auxiliary line, and was seen at a glance.

3. I incorrectly self-imposed this condition. As Polya said, the first step to problem-solving is to understand the problem. The solution for A outside the disk can use a similar angle-based argument—it's slightly different, so I'll take some time to post it for other readers later.

1. What does N have to do with anything? You need to show that AM < AP.

2. Your argument to show that AM > AQ works, and it is also possible to show that AM<AP by the same argument, based on angles. [Why do you use different arguments for proving similar statements?]

Yet, what if A is outside the circle? (Your argument based on angles should also work, but might appear different due to a new disposition of the points involved).

Hey, thanks. I was intentionally being a bit informal; didn't want to spoil it. This is the argument I gave in my notebook:

Let a circle with centre O be given, and an arbitrary point A in its interior region also be given. Draw a secant through O and A, and mark its intersections with the circle as Q and P. Choose an arbitrary point M along the circle not equal to P or Q, and draw a line from it through A up to its intersection with the circle at N.

1. (part 48) NA < OA + ON, and ON = PO as radii of the same circle. Therefore, NA < OA + PO, and thus, because PO + OA are summands of the segment PA, and N is arbitrary, P is the farthest point from A.

2. Draw a line from M to O, then OM = OQ as radii of the same circle, so the triangle OQM is isosceles, and thus (part 35) the angles at the base are congruent (i.e. OQM = OMQ). But then in the triangle AQM, the angle AMQ < (AQM = OQM) because AMQ is a part of angle OMQ. Therefore, (part 45) AM > AQ as a side opposite to a greater angle in a triangle. Because M was chosen to be arbitrary, Q is the closest point on the circle to A.

It doesn't sound like my argument for point 2 was what you had in mind. Seemed a bit easy for an asterisked question. I'll see if I can approach the problem from another perspective. If it's possible to boost my karma, then I would appreciate it if you would do that so that I can post images here. Thanks.

Hmm, I don't think you gave a valid solution. First it is unclear: Which angle is smaller than which? What is "angle between" two points? Second, the "given point" is not required to be interior - it could be exterior to the circle. Third, I doubt you can get away with arguments based on comparison of angles: At this point in the text, nothing is known about angles relative to their position to a circle, so it would be hard to justify any claim, even when it is valid. Fourth, "easily determined" is not an argument: "the farthest point is easily determined via triangle inequality" is merely a rephrasing of the hint given to the problem.

I believe, if you make an accurate argument (based on this hint) for the farthest point, you'll see that a similar argument works for the closest point as well. Good luck!

I'd like to upload a photo, but I can't. The farthest point is easily determined via the triangle inequality. For the closest point, take an arbitrary point on the circle, and join it to the centre. As two radii, the distance of the 'closer' point on the secant is the same as that of the arbitrary point's to the centre. The triangle formed by the arbitrary point and the 'closer' point along with the centre is therefore isosceles. Join the arbitrary point on the circle with the given interior point, and the angle formed will be smaller than the angle between the 'closer' point and the centre. The solution follows from here by part 45.

arbitrary point*. The assumption that M and N concurrently cannot be taken is also unnecessary.

Exercise 219: Given an angle and a point inside of it, construct a triangle with the shortest perimeter such that one of its vertices is the given point, and the other two lie on the sides of the angle.

Solution:

Let angle BAC be given, and a point Q in its interior also be given. Mark the point D symmetric to Q about AB, and mark the point E symmetric to Q about AC. Draw a line from D to E, and mark the points where it intersects AB and AC as M and N, respectively. Draw lines from Q to both M and N, and from M to N. The triangle QMN is the required construction, ie. that triangle which satisfies the given conditions of the problem.

Proof: Take an arbitrary on each side of angle BAC except both M and N concurrently, and mark the one on AB as J, and the one on AC as K. Join J to D and K to E by lines. Notice that QJK = DJ + JK + KE (by the properties of axial symmetry), and this broken line is smallest when it is straight (part 49), ie. DM + MN + NE = QMN.

Alternative (and possibly simpler): Let two lines PQ and MM parallel to each other be given, and a point O contained between the region of the plane bounded by PQ and MM also be given. Let the segment to be constructed, a, be given.

(1) If a is less than the distance from PQ to MM (as defined by part 85), then there is no solution because any perpendicular dropped from PQ to MM (or vice-versa) would be smaller than any slant (part 51), and a will only form a part of the perpendicular.

(2) If a is not less than the distance from PQ to MM, then it can be constructed through O, with endpoints on PQ and MM, respectively, using the following process:

Choose a point A along MN, and setting the compass to a step equal to a, place the pin leg at A. Using the compass' pencil leg, describe an arc of radius a, and mark its intersection with PQ as B.

Draw a line from A to B. If O lies along A and B, then we are done, so suppose it does not. Then, through the point O, construct a line parallel to AB, and mark its intersections with PQ and MN as C and D, respectively.

ABCD is a parallelogram (because by hypothesis, PQ || MN, and this implies CB || DA as segments of respectively parallel lines, and CD || AB by construction), so by part 85, CD = AB = a as opposite sides of a parallelogram.

But O belongs to CD by construction, and so we have constructed the required segment.

perpendicular to AB or its extension in either direction*

Erect a perpendicular to AB at C, and mark the segment CD congruent to h. Draw a line parallel to AB through D. By (85), all of the points satisfy the condition. Extending CD past C, and marking the segment CD' = h, and similarly constructing a line parallel to AB through D' will also yield a line whose points all satisfy the condition. It is now required to show that any such segment perpendicular to AB and congruent to h belongs to one of the two parallel lines; I omit this part because there is likely a more natural solution than what I have conceived.

Thanks!

Thanks.

It's taught using the first book published by the AMS. The course (Math 223: Linear Algebra) is taught at UBC Vancouver, and it's taught by Jozsef Solymosi.

The answer to your question is "no"; I am not sure which linear algebra book you are talking about (there are two), but none of them assumes any systematic exposure to elementary geometry (although probably assumes some common-sense intuition about it). The book "Linear Algebra and Differential Equations" published by AMS corresponds to an honors version of a sophomore level course Math 54 taught at UC Berkeley. For a regular, non-honors Math 54, it is too terse. (So, if this is the book, I am curious where and who teaches the course using it.) The other book is unpublished but is currently available online in several drafts (the latest is found
here and corresponds to a junior-level, upper-division course in Linear Algebra. It is formally independent of the first one, and at the beginning contains a couple of sections about vectors in elementary geometry essentially copied from my adaptation of Kiselev's "Stereometry".

Can a student tackle your (in development) book for a linear algebra course if he does not have any command of solid geometry?

I'm considering registering for a course which teaches using your book, but my only meaningful contact with geometry will have been through the (expected by that time) completion of Kiselev's Geometry Book I. Planimetry. Should I postpone the course until I've completed Book II. Stereometry? Because of the existence of your translations of Kiselev's series about geometry, I suspect that a course based on your writings might come with some high expectations (concerning Euclidian geometry); without meeting those expectations, I suspect that a student cannot master the material.

Thank you.

"and hence it is also the center of symmetry of GFJH (in notations of your figure), which is therefore a parallelogram."

This can be interpreted as saying that the diagonals bisect each other, which (by part 87 (2)) means that the figure is a parallelogram.

First, you misquoted the problem: it asks to draw a line such that the segment of it contained between two given lines is bisected by the given point, but the given lines are not said to be parallel. Second, you are right of course that if the given point lies on one of the lines, or if the lines are parallel, than the solution might not exist. However, finding the situations when there is no solution, and finding the number of solutions in each case is, strictly speaking, a part of solving a construction problem.

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