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Previous Puzzles of the Month + Solutions

September-October 2008, Puzzle nr 119
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Puzzle # 119 Difficulty level: bulbbulbbulb, general math knowledge.


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.


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
= 90° - α

b) According to the drawing:
A’C’ = AC’
MA = MA' = R - r
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)

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)
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
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.


© 2005 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.

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


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