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# Help in geometry!

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A square  varies inside equilateral triangle  so that  always lies on side   always lies on side  and  always lies on side  The point  starts on side  and ends on side  The diagram below shows the initial position of square  an intermediate position, and the final position.

Show that as square  varies, the height of point  above  remains constant.

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

Apr 12, 2018

#1
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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))."

Apr 12, 2018

#1
+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))."