# A rectangle, given area and angle

$$ABCD$$ 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 $$ABCD$$.

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!

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.

• Thank you for the response! Can you clarify how from $\dfrac{a^2+b^2}{2}=a\cdot(a+b)$ you got $b=(\sqrt2+1)\cdot a$? Commented Jan 30, 2022 at 19:10
• @Medi we have $a^2 + b^2 - 2ab = 2 a^2$ or $(b-a)^2 = 2a^2$ and we get $b = a \pm a \sqrt2$ Commented Jan 30, 2022 at 19:15
• Thank you! Why are we assuming that $b>a$? Commented Jan 30, 2022 at 19:28
• 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? Commented Jan 30, 2022 at 19:59
• I got it by myself! Thank you! Be safe! Commented Jan 30, 2022 at 20:23