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$ABCD$ is a rectangle with area $S$ and $AC\cap BD=O$. The circumscribed circle of $\triangle ABO$ intersects for second time the line $AD$ at $M$, such that $\tan\measuredangle ABM=1$. Find the diagonal and the perimeter of the rectangle $ABCD$.

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I think that we can say $\measuredangle ABM$ is an acute angle because it's one of the angles in the right-angled triangle $ABM$. Is this so? This means $\measuredangle ABM=45^\circ$ or $AB=AM$. So the length of the segment $DM$ is actually the sum of the sides of the rectangle $ABCD$.

It is worth noting that the centre of the circumscribed circle of $\triangle ABO$ lies on $BM$ because $ABM$ is also inscribed in that same circle and $\measuredangle BAM=90^\circ$. How does the given area $S$ come to play? Thank you!

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Say $AD = a, AB = b$. Then $ab = S$

We also know $AM = AB = b$, given $\angle ABM = 45^\circ$

Using power of point $D$,

$OD \cdot BD = AD \cdot MD$

Or, $ \displaystyle \frac{BD^2}{2} = \frac{a^2+b^2}{2} = a \cdot (a+b)$

And we get, $b = (\sqrt2 + 1) \cdot a$

Now knowing relationship between $a$ and $b$ and also knowing that $ab = S$. Can you find $a$ and $b$ in terms of $S$? From there, you can find the diagonal and the perimeter.

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  • $\begingroup$ Thank you for the response! Can you clarify how from $\dfrac{a^2+b^2}{2}=a\cdot(a+b)$ you got $b=(\sqrt2+1)\cdot a$? $\endgroup$ Commented Jan 30, 2022 at 19:10
  • $\begingroup$ @Medi we have $a^2 + b^2 - 2ab = 2 a^2$ or $(b-a)^2 = 2a^2$ and we get $b = a \pm a \sqrt2$ $\endgroup$ Commented Jan 30, 2022 at 19:15
  • $\begingroup$ Thank you! Why are we assuming that $b>a$? $\endgroup$ Commented Jan 30, 2022 at 19:28
  • $\begingroup$ So for the perimeter I got that $P=2(a+b)=2\left(\sqrt{S(\sqrt2+1)}+\sqrt{S(\sqrt2-1)}\right)$. The given answer is $2\sqrt{2S(1+\sqrt2)}$. Am I wrong in the way? $\endgroup$ Commented Jan 30, 2022 at 19:59
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    $\begingroup$ I got it by myself! Thank you! Be safe! $\endgroup$ Commented Jan 30, 2022 at 20:23

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