Euclidean and non Euclidean geometries problems

AI Thread Summary
The discussion revolves around verifying solutions to problems from the book "Euclidean and non Euclidean geometries" by Greenberg. The user presents several answers to geometry problems, seeking confirmation on their correctness. Responses indicate that asking for verification of multiple answers in one post is less effective, as it may discourage engagement. Participants suggest focusing on individual questions for better feedback and note that the forum is not primarily for homework checking. The overall sentiment emphasizes the importance of concise inquiries for fruitful discussions.
nmego12345
Messages
21
Reaction score
0
So I was reading this book, "Euclidean and non Euclidean geometries" by Greenberg
I solved the first problems of the first chapter, and I would like to verify my solutions


1. Homework Statement

UJAOmle.png


iEmgmcx.png


Homework Equations


[/B]
Um, none that I can think of?

The Attempt at a Solution



(1) Correct
(2) It is defined as: "An "angle with vertex A" is a point A together with two distinct nonopposite rays AB and AC (called the sides of the angle) emanating from A
We're asked if it is defined as the space between two rays that emanate from a common point
I think that's incorrect
(3) Incorrect
(4) Incorrect (2 lines are parallel if they don't intersect)
(5) Incorrect (it wasn't proven)
(6) Correct
(7) Incorrect (it is a right angle if it has a supplementary angle to it to which it is congruent
(8) Correct
(9) Correct
(10) Correct

Definitions:
1. (a) Midpoint M of a segment AB
M is the midpoint of the segment AB if M lies on segment AB where the segment MA is congruent to the segment MB
1. (b) Perpendicular bisector of a segment AB
l is the Perpendicular bisector of the segment AB if the line l is perpendicular to line AB (we've already defined that two lines (l, m) are perpendicular if they intersect at a point A and if there is a ray AC that is part of l and a ray AC that is part of M such that angle BAC is right angle) and if the point of intersection of the lines l and AB is also the midpoint of the segment AB
1. (c) Ray BD bisects angle ABC if angle ABD is congruent to the angle CBD
1. (d) Points A, B, and C are collinear, if the rays BA and BC are opposite
1. (e) lines l, m and n are concurrent if the lines l and m intersect at a point A and the lines m and n intersect at point A as well

2.(a) The triangle ABC formed by three collinear points A, B and C is the set of points that lie on the segment AB + the set of points that lie on the segment BC + the set of points that lie on the segment CA
2.(b) The vertices of ABC are the three points that lie on the triangle ABC in which if we draw line segments joining the 3 of them, we'll get a triangle congruent to triangle ABC
The sides of ABC are the 3 segments that lies on the triangle ABC where the first and second segment intersect at the first vertex, the second and third segment intersect at the second vertex, and the third and first segment intersect at the third vertex
The angles of ABC are the 3 angles in which each angle is formed of the union of a vertex of tirangle and the 2 rays that emanate from that vertex in which the two sides that intersect at the vertex each of them is part of one of the 2 rays.
(Should I say that they are called angles ABC, CAB and ACB or I don't have to?)
I want to safecheck my solutions before solving other problems
 
Last edited:
Physics news on Phys.org
Anyone, help?
 
I guess it is hopeless
 
No one is going to read a post this long just to confirm it for you. You've asked 10 questions in one post!
This is something you can do with your friends at university.
 
nmego12345 said:
I guess it is hopeless

carpenoctem said:
No one is going to read a post this long just to confirm it for you. You've asked 10 questions in one post!
This is something you can do with your friends at university.

Indeed. Asking for people to check your answers is unlikely to get many responses. You're far more likely to get a response if you post a single question that you're having trouble with and want to talk about. PF isn't really a homework-checking site. There are plenty of those elsewhere.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top