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A square $DEFG$ varies inside equilateral triangle $ABC,$ so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overline{BC},$ and $G$ always lies on side $\overline{AC}.$ The point $D$ starts on side $\overline{AB},$ and ends on side $\overline{AC}.$ The diagram below shows the initial position of square $DEFG,$ an intermediate position, and the final position.

[asy] unitsize(3 cm);pair A, B, C, D, E, F, G, trans = (1.4,0);A = dir(60);B = (0,0);C = (1,0);D = interp(B,A,(1 + 1/sqrt(3))/(1 + 2/sqrt(3)));E = interp(B,A,(1/sqrt(3))/(1 + 2/sqrt(3)));F = interp(B,C,(2/sqrt(3))/(1 + 2/sqrt(3)));G = interp(C,A,(1)/(1 + 2/sqrt(3)));draw(E--F--G--D--cycle,red);draw(A--B--C--cycle);label(

Show that as square $DEFG$ varies, the height of point $D$ above $\overline{BC}$ remains constant.

[asy] unitsize(4 cm);pair A, B, C, D, E, F, G, P;A = dir(60);B = (0,0);C = (1,0);F = (0.52,0);E = extension(A,B,rotate(90,F)*(A),rotate(90,F)*(C));G = rotate(-90,F)*(E);D = E + G - F;P = (D + reflect(B,C)*(D))/2;draw(D--E--F--G--cycle,red);draw(A--B--C--cycle);draw(D--P,dashed);label(

 

 

Please explain in detail, I reallyyy don't get this problem :(

Guest Apr 12, 2018

Best Answer 

 #1
avatar+8 
+1

I was looking around and saw that this question has been solved before. The previous answer stated

"Begin by moving F a little to the left so that the square has begun it's rotation.

Call the angle EFB theta, call the side of the square b, and BF and FC x and y respectively.

Fill in the angles of the two triangles BEF and FGC and apply the sine rule in both triangles.

Use the fact that x + y = a, the length of the side of the equilateral triangle to deduce that

.

So the size of the square varies as it rotates.

Now look at your larger diagram, the one with the dotted line running from D.

Call the point where this line meets BC N, then  DN/DF = sin(angle NFD).

DF  is the diagonal of the square = b.sqrt(2) and angle NFD = theta + 45 deg, so you can now work out DN.

You should find that it's a constant, equal to  a.sqrt(3)/(1 + sqrt(3))." 

Made by Guest.

Pheonix  Apr 12, 2018
 #1
avatar+8 
+1
Best Answer

I was looking around and saw that this question has been solved before. The previous answer stated

"Begin by moving F a little to the left so that the square has begun it's rotation.

Call the angle EFB theta, call the side of the square b, and BF and FC x and y respectively.

Fill in the angles of the two triangles BEF and FGC and apply the sine rule in both triangles.

Use the fact that x + y = a, the length of the side of the equilateral triangle to deduce that

.

So the size of the square varies as it rotates.

Now look at your larger diagram, the one with the dotted line running from D.

Call the point where this line meets BC N, then  DN/DF = sin(angle NFD).

DF  is the diagonal of the square = b.sqrt(2) and angle NFD = theta + 45 deg, so you can now work out DN.

You should find that it's a constant, equal to  a.sqrt(3)/(1 + sqrt(3))." 

Made by Guest.

Pheonix  Apr 12, 2018

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