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The sides $AB, BC, CD$ of trapezoid $ABCD$ touches the circle with center $O$ and they are equal. $AD$, goes through the point O. If diameter is 2, then the area of the trapezoid is $√a$ . What is the value of a?

Source: Bangladesh Math Olympiad 2016 Junior Catagory

I tried but I could not find any possible solution. How am I supposed to get $BC + AD$?

  • 1
    $\begingroup$ Is it not just half of the regular circumscribed hexagon? $\endgroup$ – dfnu Jan 21 at 17:02
  • $\begingroup$ Out of curiosity, do you have the actual answer? Maybe not the solution, but an answer at least? I ask because I think I might have a solution but I'd like to double-check because it feels a little convoluted. $\endgroup$ – Eevee Trainer Jan 21 at 17:02
  • $\begingroup$ First show that $ABO$, $BOC$, and $OCD$ are equilateral triangles... $\endgroup$ – dfnu Jan 21 at 17:21

Warning: might be a little convoluted, uses a bit of assumed knowledge, and is probably not the most elegant/efficient solution. I've triple-checked the method and arithmetic though so unless there's a fundamental flaw with my solution, I think this should be correct.

We make a duplicate trapezoid of $ABCD$, with diameter $AD$, which basically inscribes the circle in a regular hexagon:

enter image description here

Since the duplicate trapezoid has the same area, the area of the hexagon is $2\sqrt a$.

The area of a regular hexagon can be shown to be given by $\frac{3 \sqrt 3}{2} s^2$, where $s$ is the length of a side of the hexagon. Thus,

$$2 \sqrt a = \frac{3 \sqrt 3}{2} s^2 \implies a = \frac{27}{16} s^4$$

The area of a regular polygon can also be shown (link above) to be given by $xp/2$, for perimeter $p$ and apothem $x$. Here, $p = 6s$, and $x = 1$, the radius of the circle, clear by the construction above and that the diameter is $2$. Thus, the area is given by $(6s)(1)/2 = 3s$.


$$2 \sqrt a = 3s \implies a = \frac{9}{4}s^2$$


$$a = \frac{27}{16}s^4 = \frac{9}{4}s^2$$

We subtract the right fraction from the left, and multiply through by $16$ as we begin solving for $s$. We get the equation

$$27s^4 - 36s^2 = 0$$

Let $u = s^2$ for ease of use, then the above equation becomes

$$27u^2 - 36u = 0 \implies 9u(3u - 4) = 0 \implies u = 0, u = \frac{4}{3}$$

$u=0$ is obviously not what we want, but $u=4/3$ is fine. Then, going back through our substitutions,

$$u=\frac{4}{3} \implies s^2 = \frac{4}{3} \implies s = \frac{2}{\sqrt 3} \implies 2 \sqrt a = 3 \cdot \frac{2}{\sqrt 3} \implies a = 3$$

  • $\begingroup$ Just add up three areas of equilateral triangles with altitude equal to the circumference radius and get the same result much quicklier. $\endgroup$ – dfnu Jan 21 at 17:42


Once you showed that $ABO$, $OBC$ and $COD$ are equilater triangles, note that the altitude of such triangles is equal to $r=1$. By Pythagorean theorem on half of the triangle determine the side, i.e. $\frac{2\sqrt{3}}{3}$. The total area of the trapezoid is equal to three times the area of the triangle, i.e. $\sqrt{3}$


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