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# Geometry Puzzle: Difficulty level 2/5

Question asked by Kaustubh Katdare in #Brainy Puzzles on May 12, 2009 Kaustubh Katdare · May 12, 2009
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Here's a geometry puzzle for you.

Source: https://www.archimedes-lab.org/

[FONT=Verdana, Arial, Helvetica, sans-serif]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 below), try to guess the area of the blue square WITHOUT using trigonometry! [/FONT] Posted in: #Brainy Puzzles RajdeepCE · May 16, 2009
The area of square is perhaps "20.25 cm square ", cause the length of square should be 4.5 cm.
Am I right?
I just guessed the answer by looking at diagram. But how can we solve this without using trignometry? Can you show us the solution with proper method, without using trignometry?!!! shalini_goel14 · May 16, 2009
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Ya! I also guess it might be 20.25 cm[sup]2[/sup]. Kaustubh Katdare · May 16, 2009
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Could you all explain how you arrived at the answer? 😀 RajdeepCE · May 16, 2009
I thought that without using trignometry, its almost impossible to find that. So I go straight, took my scale, and measure the length!!!! The length of square is about 4.5 cms, so I guessed that may be this should be the right answer.
If this answer is wrong, then show us the answer and ofcource with explanation how to do that without using trignometry. Anil Jain · May 17, 2009
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RajdeepCE
I thought that without using trignometry, its almost impossible to find that. So I go straight, took my scale, and measure the length!!!! The length of square is about 4.5 cms, so I guessed that may be this should be the right answer.
If this answer is wrong, then show us the answer and ofcource with explanation how to do that without using trignometry.
I do not know answer is correct or not but I am sorry to say Rajdeep, as per me this is the least preferred way a Engineer would like to solve this problem.

Apply some logic or geometry theorum to solve this puzzle.

-CB RajdeepCE · May 17, 2009
You are right Crazyboy. I typed this answer cause I want to add humours cause none is replying to this puzzle. And I thought it would have fun, but I was wrong.
Anyway I just guessed the answered by just looking at figure. And one more thing about this problem, this problem is may be on lateral thinking cause may be puzzler want to check that how we respond to puzzle. And last but not the least, I thought its impossible to find the answer without trignometry. Now I am just waiting for The_big_K to post the correct solution. Anil Jain · May 17, 2009
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Ok RajDeep, thats fine.

@ Shailini - Any explaination what make you think that answer can be 20.25 cm2.

For me, Answer would be somewhere around 15-16 cm2. Reason being, diameter of big circle is 10 so radius is 5 and area of half bigger circle is 26.75. Area of quarter circle is 13.375 cm2. Definately area of square will be much lesser then 20 or I will say it would be around 15 (near to 13.375). I wish I would have value of Q, so that I could give answer of this question.

-CB sanandsanu · May 22, 2009
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i think the answer is 12.5 Kaustubh Katdare · May 22, 2009
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sanandsanu
i think the answer is 12.5 RajdeepCE · May 27, 2009
Hey biggie, can we have the answer of this puzzle, cause its have been long enough!!!!
Hope you will not disappoint us. sanandsanu · May 27, 2009
Rank E1 - BEGINNER
RajdeepCE
Hey biggie, can we have the answer of this puzzle, cause its have been long enough!!!!
Hope you will not disappoint us.
i am not sure whether it is right or not...i just guessed the answer by looking at the figure...i think the diagonal of the square is equal to the radius of the semi circle and area of a square is the square of its diagonal divided by two..hence the area should be 12.5..in this way we don't have to use trigonometry also..i am not sure if it is correct or not because i don't know how to prove that the diagonal of the square is equal to the radius Kaustubh Katdare · May 27, 2009
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I'm waiting for the author to release the solution. I'll post it as soon as it's available. skipper · Jun 5, 2009
Rank C2 - EXPERT This puzzle is an illustration of a slice group symmetry.
Q, the radius of the yellow circle, is 'outside' the circle and the square, P.
This is a numerical map of the rationals, over a section of Pi, or how quadrature of circle Y, fails when Q - P is beyond a/2 the outer 'products' diameter for Q.

And Kepler's golden ratio, since the trigonometry is, when you do use it it shows that a Kepler distance is the root of any q, along the intersection of P, for radius Q. Q is 'on' X where the tangent of Y is.
The letters have not been chosen arbitrarily.
I might be pushing my feeble grasp of symmetry groups, but I think Q0 is the centroid, and P is a helicoid over S3 the sphere, the slice group is this map of the circle U(1) subgroup with the intersection (slice) of P through Q. Q = Y, and X is all tangents on Y and roots in Q on P.

Could you 'not use' trigonometry by using a ruler to extend Q to the tangent on the outer semicircle, radius a/2? Also produce line P' from this new point, a reflection of Q, i.e. from Q' to the lower tangent on a/2? Quadrature is in the a/2 upper half plane of left-right, - +, + +, quadrants, divided at Y = 0 = a - a/2.
Then Q' is less than any X,X' values for P, along -a/2. When P = a/2, the blue square is larger in area than Y, and Q' is the (invariant/variant) radius of Y. The triangle QQ'X, where X is the P' intersection, is the eccentricity for (Q,P) the Archimedes spiroid over X,Y.
Or something. Might need a bit more gold.

Hm, note also that as it's drawn, P doesn't slice Y enough that if Q rotates Y, P will section Y completely - it isn't deep enough to be a 'deep-slice' puzzle, ala Rubik's group on the cube. So until P approaches Q and the centroid, no cube symmetry groups appear since Y is only shallow-cut.

Now consult inertial frames of reference on plane of ecliptic for planet earth that extend to 'infinity', or to a fixed set of celestial objects called quasars and binary neutron stars, so that VLBI astronomy measures nutational changes in the earth's tilt angle along QQ' as it precesses in the total solar moment P(m) over Q, for mE the energy of the geogyroscopic helicoid, relative to at least P' the lunar moment, and S' the solar moment. Use Kepler distance-measure again to verify Archimedes spiroid exists in planetery motion, up to 0.001(pi)C the circular measure of e in the rationals. skipper · Jun 6, 2009
Rank C2 - EXPERT
Another comment: "not using trigonometry" is a bit of a cheat since you have to use a protractor and ruler to draw the diagram. Has anyone tried this, and how did you do it, what algorithm was applied?
How did you determine where the square and circle should be located?

After some trial and error, I finished copying the diagram and added some more lines. I think the angles from X' to Q', where X' = intersection of vertical 'leading' edge of P with a; the angle Q makes with Q'; and the angle X makes with Q', where X is the tangent of circle Y, on a, vary by "exactly" 10 degrees. Starting 60, 70, then 80 degrees respectively.
The symmetry is this equality between the tangents on Y, and intesections P, on a and a/2.
Another line from the point P, on Y, to X, describes a 60:60:60 equiangular isolateral triangle. The area is a root for Y,P and Q on a. skipper · Jun 7, 2009
Rank C2 - EXPERT
This is a Bernoulli cycloid from National Curvebank. The above diagram is a 'stuck' cycle. The curve in the below diagram from O to P and through the circle (Y), is derivable from the first, but you need a parametric formula and trig angles to plot from O at the left to O' at right, through P and Y.

The VLBI physics is related since signals from quasars and other stable objects 'slice' the earth's surface between two radio dishes, at say P,P' along the equator, or at different latitudes which are known distances from the ecliptic plane of rotation.
Using two dishes to monitor signals at points on the equator opposite each other (a/2 apart) slices the earth into two hemispheres. This curve is part of the brachistochrone "problem of least descent" for inertial bodies.
https://curvebank.calstatela.edu/brach2/brach2.htm

p.s. dang it, I made a mistake with this:
Another line from the point P, on Y, to X, describes a 60:60:60 equiangular isolateral triangle. The area is a root for Y,P and Q on a.
The triangle isn't equiangular, it can't be since the 60,70,80 angles are different to the angle from P to X (the tangent of the x axis on the Y circle).
Perhaps this is an asymmetry or skewed symmetry, that represents an irrational relation between the square at P and the Y circle?

Been playing around with some algebra and trying to derive the area of the circle in terms of the angles and sections on a = X,X'. The outer semicircle is related to the inner one by some algebraic formula that I think I've seen before. I'm using T = 4/Y as the triangle-side to circle-diameter subst. and trying to derive the side of square PX = - PX', where QX = -QX' and Q is the intersect outside Y the circle on the "X axis".

I grobbed this thread because I'm doing some stuff on triangle groups and Kepler, Bernoulli, Newton, etc. skipper · Jun 8, 2009
Rank C2 - EXPERT
Oh crap, I just figured out the distance to P from a = x axis, is 4/5 of a/2.

The tangent from (0.P) to (0,0) on Y, is root(41) = hypotenuse..
The roots of this 4,5,{0,P} triangle are then, 2, root(5), and root(41), the golden ratio is in the difference between a/2 and the height of P.

Then P = 2/5 of a; hypotenuse = tangent on Y to (0,0) from the left edge of P is root(41)/5; and where P meets a = -X' to (0,0) is "1"; so that a/2 = P + 1.

(more mistakes corrected)

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