The aim of this puzzle is to find the angle marked '?'. Answers must be in the format of exact values.

You will have to assume some basic geometric constraints. e.g. points which look like they are touching a line, are touching the line, and if lines look perpendicular, they are. Arcs create quarter circles.

Sydney Rectangle


closed as off-topic by Ankoganit, boboquack, greenturtle3141, Beastly Gerbil, dcfyj Jul 24 '17 at 18:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Ankoganit, boboquack, greenturtle3141, Beastly Gerbil, dcfyj
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Can trigonometric expressions / formulae be used ? Pls. clarify, @Jonathan Fowler $\endgroup$ – Mea Culpa Nay Jul 24 '17 at 15:06
  • 1
    $\begingroup$ @MeaCulpaNay Yes. You can use any means you wish. The solution I used does not require trig. It does use a well-known formula though. You might need a small amount of background in maths to see it. $\endgroup$ – Jonathan Fowler Jul 24 '17 at 15:18

Begin by assuming the height of the rectangle is $1$ and the width is $1+x$.

You can then observe that there could be a rectangle between the upper right corner of the big rectangle and the peak of the smaller arc that has the same height to width ratio of the larger one.

This smaller rectangle will then have a height of $1-x$ and a width of $x$, and we can relate the two by the equality $\frac{1}{1+x} = \frac{1-x}{x}$, which can be re-formatted to: $0=\frac{1-x-x^2}{x}$

we then need to find the roots of this equation to solve for x. If x is 0, we get infinity so there's no need to consider that possibility, and we can simplify to $0=1-x-x^2$. For this we can use the quadratic formula to solve for x: $-\frac{1±\sqrt{5}}{2}$, which yields -1.618 and .618 (woot for golden ratio :D)

Obviously a negative value for x does not make sense, so we take $x=.618$. Moving to trig for a moment we can find the angle measure in question now knowing the side lengths of the larger rectangle to be $1$ and $1+.618$. This makes the expression to solve the question: $arctan(\frac{1+.618}{1})$ or: $arctan(φ)$ which is equal to 1.01722197 rad

I'm not sure if I can eliminate trig all together to get a proper angle measure out of this one. Maybe it's a common ratio I'm supposed to remember like the 30-60-90 triangle?

  • $\begingroup$ Well done. Interesting use of a smaller rectangle at the top-right. The method I used was to look at the triangle at the bottom-left, between the small arc and the base. let R be the radius of the large arc and r the radius of the smaller. The gradient of its hypotenuse is r/R. Now look at the similar triangle spanning the entire bottom-right half of the rectangle. It's hypotenuse is collinear with the previous one and it's gradient can be expressed as R/(R+r). => r/R=R/(R+r) and from there it is identical to your method. $\endgroup$ – Jonathan Fowler Jul 24 '17 at 17:01
  • $\begingroup$ @Aaron, not all all rectangles are eligible to obtain the so called 'Golden Ratio'. Please check. Accordingly the question 'may' become inapt ! $\endgroup$ – Mea Culpa Nay Jul 25 '17 at 6:59
  • $\begingroup$ @MeaCulpaNay I don't believe I have made any mathematical errors, but if you have found one, please point it out. $\endgroup$ – Aaron Jul 25 '17 at 14:03
  • $\begingroup$ @Aaron, on what basis, the below said 'Rectangles' are similar ? "You can then observe ... there could be a rectangle between the upper right corner of the big rectangle and the peak of the smaller arc that has the same height to width ratio of the larger one". As such there is NO theorem for similarity of rectangles, as I know. If you can prove by establishing any facts, please provide the required explanation. Thanks. $\endgroup$ – Mea Culpa Nay Jul 25 '17 at 14:57
  • $\begingroup$ @JonathanFowler, can you please explain how you solve further, after establishing the equation - r/R=R/(R+r) ? It is not clear for me. Thanks. $\endgroup$ – Mea Culpa Nay Jul 25 '17 at 15:05

An attempt:

Let radius of the bigger arc be $\text{R}$ and that of smaller arc be $r$. And let $\frac{\text{R}}{r} = k$.
Then we have $\sin(?)=\frac{(\text{R}+r)}{\sqrt{(\text{R}+r)^2 + r^2}}$.
Based on the value of $k$, it can be shown that the angle is — $\sin^{-1} \frac{(1+k)}{\sqrt{(1+k)^2 + k^2}}$.

Too much of algebra...hmmm.. though the question is of geometric nature!

  • $\begingroup$ "Beneath a geometry problem lies a tedious algebra exercise" (William), therefore, "Good luck!" (Taken) $\endgroup$ – William Nathanael Jul 24 '17 at 15:18
  • $\begingroup$ Do you mean Sin (?) = (R+r) / Square Root ((R+r)^2 + (R)^2)? When you get an answer, it won't need to be expressed in terms of any variables, like 'k'. e.g. an answer might look like asin(sqrt(2)-1). Keep at it. I'm a maths lover, so it does have some calcs needed. I personally enjoy that sort of thing. $\endgroup$ – Jonathan Fowler Jul 24 '17 at 15:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.