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Previous Puzzles of the Month + Solutions

 
May-June 2009, Puzzle nr 121
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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.


solution

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

SOLUTION 1

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
QR
ON
QPOP
NQO = ∠OQP

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

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

Congratulations!


© 2006 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.


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
http://mathafou.free.fr/pbg_en/jsp136e.html 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.

 



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