Question

There are two squares S₁ and S₂ with areas 8 and 9 units, respectively. S₁ is inscribed within S₂, with one corner of S₁ on each side of S₂. The corners of the smaller square divide the sides of the bigger square into two segments, one of length a and the other of length b (b > a). A possible value of b/a is:

• A. $\ge 5$ and $< 8$
• B. $\ge 8$ and $< 11$
• C. $\ge 11$ and $< 14$
• D. $\ge 14$ and $< 17$
• E. $>17$

Hint:

For a square of side $s$ rotated by $\theta$ inside a square of side $L$, use $s(\cos\theta+\sin\thet(A)=L$. The split lengths along a side are proportional to $\cos\theta$ and $\sin\theta$, so the ratio becomes $\max(\tan\theta,\cot\thet(A)$.

Answer 1

The Correct Option is D

Step 1: Relate the rotation angle to side lengths.
Let the side of the big square be \(L=3\) (Area \(9\)) and the side of the small square be \(s=\sqrt{8}=2\sqrt{2}\).
If the small square is rotated by an angle \(\theta\) (acute) from the sides of \(S_2\), then the bounding-box relation gives \[ s(\cos\theta+\sin\theta)=L ⇒ \cos\theta+\sin\theta=\frac{L}{s}=\frac{3}{2\sqrt{2}}. \] Hence \[ (\cos\theta+\sin\theta)^2=1+\sin 2\theta=\frac{9}{8} \ ⇒\ \sin 2\theta=\frac{1}{8}. \]

Step 2: Express the segment ratio in terms of \(\theta\).
Each side of \(S_2\) is split by a vertex of \(S_1\) into segments proportional to \(\cos\theta\) and \(\sin\theta\):
\[ a : b=\min(\cos\theta,\sin\theta):\max(\cos\theta,\sin\theta), \] so \[ \frac{b}{a}=\max\!\left(\tan\theta,\cot\theta\right). \] Let \(t=\tan\theta>0\). From \(\sin 2\theta=\dfrac{2t}{1+t^2}=\dfrac{1}{8}\),
\[ \frac{2t}{1+t^2}=\frac{1}{8} \ ⇒\ t^2-16t+1=0 \ ⇒\ t=8\pm 3\sqrt{7}. \] Thus \(\max(t,1/t)=8+3\sqrt{7}\approx 15.94\).

\[ \boxed{\dfrac{b}{a}=8+3\sqrt{7}\approx 15.94} \] which lies in the interval \([14,17)\).