Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Learn to enjoy mathematics.
Google
Web CTK
Best sites for teachers
Sites for teachers
Sites for parents
Terms of use
Awards

Interactive Activities
CTK Exchange
CTK Insights - a blog

Games & Puzzles
What Is What
Arithmetic/Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Eye Opener
Analog Gadgets
Inventor's Paradox
Did you know?...
Proofs
Math as Language
Things Impossible
Visual Illusions
My Logo
Math Poll
Cut The Knot!
MSET99 Talk
Other Math sites
Front Page
Movie shortcuts
Personal info
Reciprocal links
Privacy Policy

Guest book
News sites

Recommend this site

Best sites for teachers
Sites for teachers
Sites for parents

Education & Parenting

Manifesto: what CTK is about Search CTK Buying a book is a commitment to learning Table of content Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

The 80-80-20 Triangle Problem, A Variant

The isosceles triangle with the apex angle of 20° and the base angles of 80° is a recurring subject of geometric problems. The most popular is the one where cevians are drawn from the base vertices at angles of 50° and 60°, and the task is to determine one of the so formed angles.

A novel problem has been suggested by Radheyshyam Poddar:

  ABC is an isosceles triangle with vertex angle BAC = 20° and AB = AC. Draw BCD = 60°; D lying on AB. Draw an arc with B as center and radius equal to BC. Let this arc cut AC at point E and AB at the point F. Prove that CE = DF.
 

Solution

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

Solution

Below is a solution by Mariano Perez de la Cruz.

 
  1. From point B draw a line BD', a reflection of CF the bisector of the angle at A. So BD' will form obviously 60° angle with respect to base BC like is the case for CF.

  2. BD' intersect to CF in P; two of the angles of ΔBPC are 60°, the angle at P must be 60° too; so BPC is equilateral and all its side are equal: BC = BP = CP.

  3. Since the arc EP of the circle is common to the angles ECP and angle EBP and since point C lies on the circle and B at its center, the central angle EBP equals twice the value of angle ECP which is 40° so the ΔBEF is equilateral too and its side EF also equals BC. So point M of intersection of EF and CD belongs to the bisector of angle CBF.

  4. By mirroring with respect to line BM the ΔMDF we get ΔMD''C, with D'' on the extension of BC. Since angle ED''C equals 40°, angle D''EC also equals 40° which means that ΔD''CE is isosceles and side CE equals to CD'' which is the mirror image of FD. Q.E.D.

    Copyright © 1996-2008 Alexander Bogomolny

28760757Page copy protected against web site content infringement by Copyscape


Search:
Keywords:


Latest on CTK Exchange
Math
Posted by Laura
2 messages
06:56 AM, Apr-15-08

Divisibility rules - Jargon buste ...
Posted by Carolyn
2 messages
08:35 AM, Apr-04-08

drawing puzzle
Posted by martin gran
31 messages
06:53 PM, May-09-08

conway's game of life
Posted by frequency
0 messages
11:52 PM, May-12-08

Mistake on the page (an aside, Be ...
Posted by Max
4 messages
10:28 AM, Feb-28-08

A Riddle
Posted by idavis1
33 messages
06:59 AM, May-15-08

Josephus Flavius (correction)
Posted by David Turner
1 messages
09:42 AM, May-14-08