In the equilateral triangle $ABC,AB=12.$One vertex of a square is at the midpoint of the side $BC$, and the two adjacent vertices are on the other two sides of the triangle.Find the length of the side of the square.

Let $DEFG$ be the square.Let $D$ be the midpoint of the side $BC.BD=DC=6.$

Let $E$ be on the side $AC$ and $G$ be on the side $AB$ such that $AG=12-a,BG=a$ and $AE=12-b,EC=b$.

In triangle $DEC,$ applying Cosine law,

$\cos 60^\circ=\frac{1}{2}=\frac{b^2+36-DE^2}{12b}........(1)$

In triangle $DGB,$ applying Cosine law,

$\cos 60^\circ=\frac{1}{2}=\frac{a^2+36-DG^2}{12a}........(2)$

I am stuck here.

  • $\begingroup$ You must have $BG=EC$, so side length is just $\frac{6\sin 60}{\sin 75}\approx 5.38$. $\endgroup$ – almagest Apr 27 '16 at 14:04

enter image description here

1) $\triangle ADB: \angle D=90^{\circ}, DB=6, AD=\sqrt{12^2-6^2}=\sqrt{108}, AB=12$

2) $DG - $ bisector $\angle ADB$

$$DG=\sqrt2 \frac{AD \cdot DB}{AD+DB}=\sqrt2 \frac{6\sqrt3 \cdot 6}{6\sqrt3+6}=\frac{6\sqrt6}{\sqrt3+1}=3\sqrt6(\sqrt3-1)$$

  • $\begingroup$ You introduced a spurious $\sqrt6$ right at the end. $\endgroup$ – almagest Apr 27 '16 at 14:06
  • $\begingroup$ @Roman83 the question seems to require that E and F be on the triangle sides $\endgroup$ – G Cab Apr 27 '16 at 14:09
  • $\begingroup$ @almagest: Thank you! $\endgroup$ – Roman83 Apr 27 '16 at 14:19
  • $\begingroup$ I want to ask one thing ,how did you know $AD$ is perpendicular to $BC$ and why $F$ came on $AD$?@Roman83 $\endgroup$ – Vinod Kumar Punia Apr 27 '16 at 14:21
  • 1
    $\begingroup$ @GCab No. The two vertices adjacent to the vertex at $D$. Roman83's triangle is rotated! $\endgroup$ – almagest Apr 27 '16 at 14:21

Let $x$ be length of side of square.Triangles DBG and CDE are congruent because they have an angle and two corresponding sides which are equal.Implies BG=EC.In which case Angle BDG = $45^0$ and Angle BGD = $75^0$.Applying sine rule in Triangle BGD $$\frac{x}{\sin{60^0}}=\frac{6}{\sin{75^0}}$$ I hope this was helpful.

  • $\begingroup$ A neat solution. $\endgroup$ – almagest Apr 27 '16 at 14:23
  • $\begingroup$ By which similarity criteria,you proved triangles $DBG$ and $CDE$ similar.Because there are only three similarity criterias,$SSS,AAA,SAS$.@VarunKumar $\endgroup$ – Vinod Kumar Punia Apr 27 '16 at 14:41
  • $\begingroup$ @Vinod Kumar Punia he appears to have used ASS congruence, which only exists in the case of right angles (as RHS). I think it Is a mistake.. $\endgroup$ – N.S.JOHN Apr 27 '16 at 15:19

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.