MHB Construct Point N: Guide to Building a Project Plan

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Point
AI Thread Summary
The discussion outlines a geometric construction to find point N based on given points and angles. It involves using ray r through point M and point B, and establishing point P such that angle PBA equals angle ABC. A perpendicular ray l from P intersects line BC at point N'. The construction then requires finding point M' on ray r, ensuring the distance from N' to M' equals the distance from P to N'. Finally, point N is determined on line BC such that triangles M'BN' and MBN are similar, completing the construction.
Albert1
Messages
1,221
Reaction score
0

Attachments

  • construct a point N.jpg
    construct a point N.jpg
    15.1 KB · Views: 93
Last edited:
Mathematics news on Phys.org
Re: Construct a point N

Albert said:
Let $r$ be the ray which passes through $M$ and whose beginning point is $B$.

Let $P$ be a point such that $\angle PBA=\angle ABC$.
Draw a ray $l$ which is perpendicular to $AB$ and whose beginning point is $P$.

Let $l$ intersect $BC$ at $N'$.
Let Find a point $M'$ on $r$ such that $|N'M'|=|PN'|$.
Now find a point $N$ on $BC$ such that $\Delta M'BN'\sim MBN$. Then $N$ is the required point.
 
Re: Construct a point N

caffeinemachine said:
Let $r$ be the ray which passes through $M$ and whose beginning point is $B$.

Let $P$ be a point such that $\angle PBA=\angle ABC$.
Draw a ray $l$ which is perpendicular to $AB$ and whose beginning point is $P$.

Let $l$ intersect $BC$ at $N'$.
Let Find a point $M'$ on $r$ such that $|N'M'|=|PN'|$.
Now find a point $N$ on $BC$ such that $\Delta M'BN'\sim MBN$. Then $N$ is the required point.

very good solution (Clapping)
 
Re: Construct a point N

Albert said:
very good solution (Clapping)
Thanks. :)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top