Cut the knot: learn to enjoy mathematics
A math books store at a unique math study site. Learn to enjoy mathematics.
Google
Web CTK
Try our no ads browsing

Sites for teachers
Sites for parents
Terms of use
Awards

Interactive Activities
CTK Exchange
CTK Wiki Math
CTK Insights - a blog

Games & Puzzles
What Is What
Arithmetic/Algebra
Geometry
Probability
Outline Mathematics
Make an Identity
Book Reviews
Stories for Young
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
Privacy Policy

Guest book
News sites

Recommend this site

Games to relax

Tutor Match Tutoring and Homework Help

Sites for teachers
Sites for parents

Education & Parenting

Manifesto: what CTK is about Buying a book is a commitment to learning Table of content Try our no ads browsing Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

The 80-80-20 Triangle Problem, A Derivative, Solution #3

 
  ABC is an isosceles triangle with vertex angle BAC = 20° and AB = AC. Point E is on AB such that AE = BC. Find the measure of AEC.

Solution

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

The solution is based on embedding the diagram into a regular 18-gon.

 

Line CE appears to pass through vertex X of the 18-gon, with CAX = 160°. Instead of showing this, we'll join vertex X to C and prove that the intersection L of CX with AB coincides with E.

 

Let Y be the vertex of the 18-gon, midway between X and C and Z the foot of the perpendicular from A to CX, i.e. the intersection of CX with AY. Since ΔACX is isosceles, CAZ = CAX / 2 = 80°. In the right ΔALZ, LAZ = 80° - 20° = 60° meaning ΔALZ is a 30-60-90 triangle with AL = 2·AZ. In order to show that AL = BC we have to show that BC = 2·AZ or, if M is the midpoint of BC, that AZ = MC.

 

To see that this is so draw AM and consider two right triangles, ACZ and ACM. These are equal by ASA: they share side AC and both have 80° angles.

We conclude that indeed L coincides with E, from which

  ECA = LCA = CAM = 10°.

And finally, AEC = 180° - 10° - 20° = 150°.

Reference

  1. R. Honsberger, Mathematical Chestnuts from Around the World, MAA, 2001, Ch 2.

Copyright © 1996-2008 Alexander Bogomolny

30864107Page copy protected against web site content infringement by Copyscape


Search:
Keywords:



Latest on CTK Exchange
try this puzzle ?/?? + ?/?? + ?/? ...
Posted by albert1950
5 messages
12:40 PM, Nov-18-08

Help me find Hisashi ABE, Pythago ...
Posted by likesmath
2 messages
11:11 AM, Oct-06-08

Bearing problem
Posted by Liliya
3 messages
11:32 AM, Nov-27-08

What is the smallest rational sqr ...
Posted by MinusOne
4 messages
04:32 PM, Nov-27-08

Three Concurrent Circles
Posted by billmillar
2 messages
12:26 PM, Oct-28-08

A geometry problem from N.A. Court
Posted by tvarhegyi
7 messages
11:48 AM, Nov-27-08

Error in Fractal Curves and Dimen ...
Posted by miguemate22
1 messages
08:51 AM, Nov-16-08