Sitemap Sitemap
Comment Contact
Newsletter Newsletter
Store Store
Syndication Features
Gallery Gallery
Games Games



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!

Life is full of misery, loneliness, and suffering - and it's all over much too soon!
     - Woody Allen

One martini is all right, two is two many, three is not enough...
     - James Thurber

You can't trust water: Even a straight stick turns crooked in it!
     - W.C. Fields



corner top left

Previous Puzzles of the Month + Solutions

May-June 2009, Puzzle nr 121
arrow back Back to Puzzle-of-the-Month page | Home arrow home


Puzzle # 121 Difficulty level: bulbbulbbulb, general math knowledge.

square and circle puzzle

Circle vs Square
The diameter a of the large semicircle below is 10 cm long. Knowing that one of the vertices of the square meets the circumference of the small circle at point P (see diagram above), try to guess the area of the blue square WITHOUT using trigonometry!

Keywords: sangaku, tangent property, bisector theorem.

Related puzzles:
- Perpendicular or not?
- Ratio of similar triangles.

Source of the puzzle:
BrainTrainer, issue #23. © G. Sarcone.
You cannot reproduce any part of this page without prior written permission.


There are 2 possible solutions to this 'sangaku' problem:


solution 1A) Build a large square with base QR (see illustration opposite) and prolong the diagonal d to the vertex S, as illustrated.
According to the drawing:
q = a/2 - R
d = QP = a/2 + x
x = d - a/2
QR = RS = a

B) According to the ‘angle bisector theorem’: if a ray or segment bisects an angle of a triangle, then it divides the two segments on either side proportionally.
In other terms, given in the triangle QRS the bisectrix QM, then:
MS / MR = QS / QR that is
(a-m)/m = asquare root2/a
We obtain, by resolving the equation:
a/(1 + square root2) = m
m = 10/2.414 = 4.14 cm

C) The right triangles QNO and QRM are similar, then:
(a/2 + x) / R = d / R = a / m = 10 / 4.14
d = R(2.42)

D) Thank to the ‘tangent property’, we find that: a/2 + x = d
Substituting x with d - a/2 in the following Pythagorean equation:
(a/2 - R)2 = x2 + R2 = (d - a/2)2 + R2
We can also substitute d with R(2.42)
(a/2 - R)2 = (d - a/2)2 + R2 = [R(2.42) - a/2]2 + R2
and we will obtain:
(5 - R)2 = [R(2.42) - 5]2 + R2
25 - 10R + R2 = (5.86R2 - 24.2R + 25) + R2
R = 14.2/5.86 = 2.423

E) The area of the blue square is therefore:
d2/2 = [R(2.42)]2/2 = [2.423 x 2.42]2/2 = 17.2 cm2


solution 2

F) According to the drawing:
q = a/2 - R ; d = QP = a/2 + x ; x = d - a/2
QP, the diagonal d of the blue square, is tangent to point P.
OP is perpendicular to QP, then:
the angle QPS = the angle OPS = 45 degrees.

G) The triangle OMP is isosceles, then the height PS of the blue square equals: ON + PM = R + R/square root2 = R(1 + square root2)/square root2 = R(1.71)
and the diagonal QP equals: R(1 + square root2) = R(2.42)

H) Thank to the ‘tangent property’ (as argument D further above), we find that:
a/2 + x = d (the diagonal QP)
Substituting x with d - a/2 in the following Pythagorean equation:
(a/2 - R)2 = x2 + R2 = (d - a/2)2 + R2
We can also substitute d with R(2.42)
(a/2 - R)2 = (d - a/2)2 + R2 = [R(2.42) - a/2]2 + R2
and we will obtain:
(5 - R)2 = [R(2.42) - 5]2 + R2
25 - 10R + R2 = (5.86R2 - 24.2R + 25) + R2
R = 14.2/5.86 = 2.423

I) The area of the blue square is therefore:
PS2 = R(1.71)2 = (2.423 x 1.71)2 = 17.2 cm2


cup winnerThe Winners of the Puzzle of the Month are:
Nheo Le, USA USA flag
Mauro Zoffoli
, Italy italian flag
Hasan Kurt
, the Netherlands netherlands flag
Amaresh G. S.
, India indian flag


© 2006 G. Sarcone,
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.

Reactions from our visitors

--- Posted: Fri, 28 May 2010 18:12:51 by Philippe Chevanne

Dear Gianni,

I just discovered your web site from a friend. Great site, and congratulations!
Lurking a few problems there, I got at problem #76 (may-June 2009)
I think the solution given is false, as it is not stated that
the diagonal QP should be tangent to the small yellow circle.
This unstated condition is necessary to solve the puzzle, and is
used in the given solutions.
Discarding this condition, the area of the square can take ANY value.
For instance see attached picture.
To construct the figure from *any* point P, that is here any
point P on a 45° diagonal line, is an "Appolonius problem" in the
case Point(P) - Line(diameter) - Circle(large circle).
That is to construct a circle (yellow) touching the three objects
considered as three "circles":
P = circle with null radius, Line = circle with infinite radius, and given (half)circle.
This can be solved in several classical ways.
(for instance at my own web site for english version)
The result is that for any point P on the 45° line, there is a
corresponding square of area = 2QP2 = any value, and a yellow
circle through P and tangent to diameter and large half circle.
(In fact two such circles, the other being "on the left" to P)
The condition "QP tangent to yellow circle" is equivallent to
"yellow circle minimal", restricting the yellow circle to be
centered in the right quarter circle, otherwise the minimum is 0,
with P=Q.
So that the missing condition "when yellow circle is minimal"
should be added to the statement.
Best Regards.
correction sangaku 121

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

angle bisector theoremAngle bisector and proportions
If a segment bisects an angle of a triangle then it divides the two segments on either side proportionally:
CA / CP = BA / PB and
CA x PB = CP x BA

Angle bisector theorem on Wikipedia.


Previous puzzles of the month...
puzzle solver
Solved Puzzles

Jan-Feb 09:
Geometric mouse

Sept-Oct 08
perpendicular or not...

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:

May 05:
stairs to square

Febr 05:
same pieces

Sept-Oct 2004:
odd triangles

Puzzle Archive
More Puzzles

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


transparent gif
comment Send a comment recommend Recommend this page facebook Share it on FaceBook stumble it Rate it on StumbleUpon
Archimedes' Laboratory™ | How to contact us | italian flag Come contattarci | francais flag Comment nous contacter
About Us | Sponsorship | Press-clippings | Cont@ct | ©opyrights | Link2us | Sitemap
© Archimedes' Lab | Privacy & Terms | The web's best resource for puzzling and mental activities
spacer spacer corner right bottom