2
$\begingroup$

I would like to solve the problem in this picture:

this picture

with just an elementary geometric approach. I already solved with trigonometry, e.g. using the Bretschneider formula, finding that the angle $ x = 15° $. Any idea?

This is how I computed the $ x $ value using the Bretschneider formula for the area of the quadrilateral $ ABDE $ and equating to the sum of the triangles' area $ ABE + EFD + BDF $

$$\begin{cases} BC = a \\ AB = a(1/\tan(2x) - 1) \\ BD = a\sqrt{2} \\ AE = AB/\cos(2x+\pi/6) = a(1/\tan(2x) -1)/\cos(2x+\pi/6) \\ ED = a/\cos(x) \end{cases} $$

So I solved this equation with Mathematica, and the only solution that fit the problem is $ x = \pi/12 $

$ a^2/2+(a^2(1/\tan(2x) - 1)(1+\tan(x)))/2 + a^2 \tan(x)/2 = ((a\sqrt{2})^2 + \\ (a(1/\tan(2x)- 1)/\cos(2x+\pi/6))^2 - (a/\cos(x))^2 -(a(1/\tan(2x) - 1))^2)/4 \tan(\pi/2 -2x) $

I guess there is a simpler trigonometric approach, but I just wanted to try with that formula.

$\endgroup$
7
  • $\begingroup$ What is the research angle to your question? $\endgroup$
    – Wlod AA
    Commented Jan 24, 2023 at 5:47
  • $\begingroup$ @WlodAA is the angle EDF $\endgroup$
    – Ulissex
    Commented Jan 24, 2023 at 9:31
  • 1
    $\begingroup$ Calling EDF the "research angle" of this question reminds me of the old joke about "finding x" (and a student pointing out "here it is"). $\endgroup$ Commented Jan 24, 2023 at 10:45
  • $\begingroup$ @JukkaKohonen 😅 $\endgroup$
    – Ulissex
    Commented Jan 24, 2023 at 10:52
  • $\begingroup$ But seriously: Trigonometric functions are an overkill and not needed at all here. You can solve it by elementary observations of sums and differences of angles. As such the problem is quite off-topic on MO, and will likely be closed. (Perhaps you meant to post it on Math StackExchange instead?) $\endgroup$ Commented Jan 24, 2023 at 11:26

1 Answer 1

1
$\begingroup$

I do not know if this is elementary enough, though the exercise could fit in an Olympiad style easily. So here is another analytic way. In the figure, let the point $D$ be the center $(0;0)$ and assume that the sides of the square are of unit length. So you get $C(1;0)$, $B(1;1)$, $A(1;\tan(X))$, $G(\frac{1}{\tan(X)};1)$.

Here we should bound $X$ so the problem has a meaning. If $E'$ is the symmetric of $E$ with respect to $F(0;1)$, your problem is equivalent to the following equation in $X$ (which characterises the angle bisector $(DE')$ in the triangle $DFG$): $$(1)\quad\dfrac{DG}{DF}=\dfrac{E'G}{E'F}.$$ Finding the line $(EA)$ as $$y=\tan(X-\frac{\pi}{6})x+\tan(X)-\tan(X-\frac{\pi}{6}),$$ then $E\left(\dfrac{1+\tan(X-\frac{\pi}{6})-\tan(X)}{\tan(X-\frac{\pi}{6})};1\right).$ Putting this in $(1)$, squaring both sides expanding and simplifying, you get the equation $$(2\sqrt{3}+1)t^4- (2\sqrt{3}+4)t^3-4t^2+ (6\sqrt{3}-4)t-2\sqrt{3}+3=0,$$ where $t=\tan(X)$. A root is $t=\sqrt{3}$, the other roots are not acceptable from the figure condition on $X$, for example $x_E<0$ etc. (I guess this may be also made elementary).enter image description here

$\endgroup$
3
  • $\begingroup$ Thank you for your proof. I still would prefer to use just geometric considerations. I tried to post here too math.stackexchange.com/questions/4623615/… and a proof was given, but still miss something (specifically why the square side is equal to the circle radius) $\endgroup$
    – Ulissex
    Commented Jan 24, 2023 at 12:16
  • $\begingroup$ Yes i guess a short proof like that is not a complete one. There shoud be some calculation. Where the problem comes from? If may i ask. $\endgroup$
    – Toni Mhax
    Commented Jan 24, 2023 at 13:05
  • $\begingroup$ Oh well from a FB group. If you are interested you can answer on Math Stackexchange too, because I guess the question on here will be closed $\endgroup$
    – Ulissex
    Commented Jan 24, 2023 at 13:09

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