2
$\begingroup$

enter image description here

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 $BC + AD$?

$\endgroup$
  • 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
2
$\begingroup$

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 $(6s)(1)/2 = 3s$.

Thus,

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

Thus,

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

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,

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

Hint.

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. $\sqrt{3}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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