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

Please solve without triginometry if posible, cause i saw the thread on this using trig but i havent taken trig yet

Nov 8, 2018
edited by bambam89  Nov 8, 2018
edited by bambam89  Nov 8, 2018

#1
+1

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)).

Tiggsy

Nov 8, 2018
#2
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Thanks for copying and pasting from the expination i dont understand and not event copying the latex. idk what i would do without you. Can anyone else help?

bambam89  Nov 8, 2018
edited by bambam89  Nov 8, 2018
#3
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I have no idea what you are on about Bambam89.

I have struggled unsucessfully with this question before.

Tiggsy's outline is excellent.     THANKS TIGGSY   I will fill the blanks in for you a little more Bambam.

Here is the initial pic that Tiggsy has described. I do not have a lot of time now but Tiggsy told you to :

1) use the sine rule to get x in terms of theta and b

2) use the sune rule to get y in terms of theta and b

3) recognixe that  a = x+y  and hence express a in terms of theta and b

this is the outcome you should be able to get.

$$a=x+y=b(1+\frac{1}{\sqrt3})(cos\theta+sin\theta)$$

See if you can get to this point and then see if you can follow Tiggsy's instructions to finish the problem.

You can let me know how you get on if you want and maybe I can help you more if you need it.

You may need to send a note with the address to me by private message to get my attention though, otherwise i may not see your post.

Melody  Nov 9, 2018
#5
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Thanks for elaborating, and it makes a little more sense, but i dont know how to use sine and sune and any of the rules bambam89  Nov 9, 2018
#4
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See if this solution will satisfy you!  https://web2.0calc.com/questions/help_92263

Nov 9, 2018
#6
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Thanks bambam89  Nov 9, 2018
#7
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Thanks guest for showing me GYangg's solution - I had not seen it before Melody  Nov 10, 2018