Previous Puzzles of the Month + Solutions

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

SOLUTION 1

A) 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
QP ⊥ OP
∠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 = a2/a
We obtain, by resolving the equation:
a/(1 + 2) = 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/2 = R(1 + 2)/2 = R(1.71)
and the diagonal QP equals: R(1 + 2) = 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

The Winners of the Puzzle of the Month are:
Nheo Le, USA
Mauro Zoffoli, Italy
Hasan Kurt, the Netherlands
Amaresh G. S., India

Congratulations!

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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.

More Math Facts behind the puzzle

Angle 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.