Shortcuts
 
sitemap
 SiteMap
shop
 Shop
books
 Books
syndication
 Print Syndication
galleries
 Galleries
Archimedes Journal
 A'Journal
e-cards
 E-cards
games
 Games
newsletter
 N'Letter
email
 E-mail

eureka!!!
corner

•••

Related links

Puzzles workshops for schools & museums.

Editorial content and syndication puzzles for the media, editors & publishers.

Abracada-Brain Teasers and Puzzles!

Numbers, just numbers...

Have a Math question?
Ask Dr. Math!

•••

•••

SMILE!Logic smile!

If you want to build a ship
don't herd people together to collect wood
and don't assign them tasks and work,
but rather teach them to long for the
endless immensity of the sea...
     - A. Saint-Exupery

Original text:
Si tu veux construire un bateau,
ne rassemble pas des hommes pour aller chercher du bois,
préparer des outils,
répartir les taches,
alléger le travail,
mais enseigne aux gens la nostalgie de l’infini de la mer...

     - A. Saint-Exupery

•••

SPACE INVADERS
Play Space Invaders online!

 

corner top left

Previous Puzzles of the Month + Solutions

 
September-October 2008, Puzzle nr 119
arrow back Back to Puzzle-of-the-Month page | Home arrow home

 

Puzzle # 119 Difficulty level: bulbbulbbulb, general math knowledge.

circle

Perpendicular or not?
  
The large circle (O1, B) above circumscribes 2 smaller circles (with centers O2 and O3, respectively) and an isosceles triangle, whose base stands on the diameter of the large circle. As shown in the drawing, each geometric shape touches the 3 other ones. Points A, O1, O2 and B are aligned. Demonstrate that the segment O3A is perpendicular to the diameter AB.

Keywords: incircles, isosceles, perpendicular, secant-tangent.

Related puzzles:
- Ratio of similar triangles.
- Soccer ball problem.


Source of the puzzle:
Sangaku tablet of Gunma prefecture (群馬県; Gunma-ken), 1803. ©G. Sarcone.


solution

Some interesting considerations
• In fig. 1, we can see that triangles A'BC' and ABC are similar.
• In fig. 2, we can see that AC' and AE' are tangent to the circle with center O3. The segment O3A represents then the symmetry axis along which the right triangle AO3D is reflected on AE'.
What we have to prove
If we prove that angle γ (gamma, see fig. 2) of the right triangle AO3D is complementary to angle α (alpha) of the isosceles triangle AC'A', then we prove that points E and E' are identical, and consecutively the segment O3A is perpendicular to the diameter AB.

solution 1
solution 2a) It is given:
radius O1B = R (see fig. 3)
radius O2B = r
radius O3D = p
segment O3A = t
β
= 90° - α

b) According to the drawing:
A’C’ = AC’
MA = MA' = R - r
h
AA'
pAC’
O1O3 = R - p
O1A = 2r - R
O1M = O2B = r
O2O3 = r + p

c) According to the 'Secant-Tangent' theorem (see further below):
t2 = p(p + 2r)

d) According to the Pythagorean theorem:
t2 + O1A2 = O1O32
t2 + (2r - R)2 = (R - p)2

e) Solving the equations in c) and d) together, we obtain:
t = 2rsquare root[2R(R - r)] / (R + r)

and
p = 2r(R - r) / (R + r)

f) According to the Pythagorean theorem:
1) h = square root(O1C'2 - O1M2) = square root(O1B2 - O1M2) = square root(R2 - r2)
and
2) AC’2 = h2 + MA2 = (R2 - r2) + (R - r)2
AC’ = square root[2R(R + r)]

g) By solving the following equations:
1) (p / t)2 = p2 / t2 = [2r(R - r)]2/ 4r2[2R(R - r)] =
= [2r(R - r)]2 / 4rR[2r(R - r)] = (R - r) / 2R
and
2) (MA / AC')2 = MA2 / AC'2 = (R - r)2 / 2R(R + r) = (R - r) / 2R
We proved that p / t = MA / AC', then the triangles AO3D and AC'M are similar, hence angle DÂO3 = angle β = 90° - α.
In conclusion, angle γ (or DÂO3) being complementary to angle α, the segment O3A is perpendicular to the diameter AB.

(You will find here a more geometric way to solve this puzzle)

 

cup winnerThe Winners of the Puzzle of the Month are:
No winners... Yes, the puzzle was a little bit harder.

Sorry!


© 2005 G. Sarcone, www.archimedes-lab.org
You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes.

You're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!


More Math Facts behind the puzzle

secant tangent theorem 1

Secant-tangent theorem: fig. 1) c2 = a(a + b) and fig. 2) c2 = a(a + 2r)

secant tangent theorem 2The angle, formed between a tangent line and a chord, is congruent to the inscribed angle on the other side of the chord and subtended by the chord (angle α' = angle α, see fig. 3). Then, with similar triangles:
c / (a + b) = a / c

 



Previous puzzles of the month...
contents+solutions
puzzle solver
Solved Puzzles

July-Aug 08:
ratio of triangles

May-June 08:
geometry of the bees

Febr-March 08
:
parrot sequence...

Dec 07-Jan 08
:
probable birthdates?

Oct-Nov 2007
:
infinite beetle path

Aug-Sept 07:
indecisive triangle

June-July 07:
Achtung Minen!

April-May 07
:
soccer balls

Febr-March 07
:
prof Gibbus' angle

Jan 07
:
triangles to square

Aug-Sept 2006
:
balance problem

June-July 06:
squared strip

Apr-May 06:
intriguing probabilities

Febr-March 06:
cows & chickens
Dec 05-Jan 06:
red monad

Sept-Oct 2005
:
magic star

July-Aug 05:
cheese!

May 05:
stairs to square

Febr 05:
same pieces

Sept-Oct 2004:
odd triangles

June 04:
pizza's pitfalls

May 04:
Pacioli puzzle...
Puzzle Archive
More Puzzles

arrow back Back to Puzzle-of-the-Month page | Home arrow home
giflet Recommend this page 

 

transparent gif
recommend Suggest this page to a friend | facebook Follow us on Facebook | comment Report any error, misspelling or dead link
Archimedes' Laboratory™ | How to contact us
| italian flag Come contattarci | francais flag Comment nous contacter
line
About Us | Sponsorship | Press-clippings | Cont@ct | ©opyrights | Tell-a-friend | Link2us | Sitemap
© Archimedes' Lab | Privacy & Terms | The web's best resource for puzzling and mental activities
spacer spacer corner right bottom