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• Module 2 Day 14 Challenge Part 1

  Reflections of Paths in a Rectangle

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  @The-Darkin-Blade 🛤 👍

• Module 2 Day 14 Challenge Part 2

  Triangle Inequality

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  @The-Darkin-Blade This mini-question is trying to ask about what the important relationships in the picture are. In geometry problems, the information we are given is often not written out in sentences or equations, but it's implicitly hidden in the diagram. 👀 So, you have to look at the elements of the picture and notice things. This question is trying to help cultivate our ability to sense what the hidden information is, and how to use it effectively.

  In the diagram for our Challenge explanation, it indeed looks like \( \angle z = \angle y,\) and the mini-question starts off with this same picture. In this diagram, \( \angle x = \angle y = \angle z.\)

   
  temp-m2d14-ch-2-screenshot.png
   

  This mini-question is asking about what influences the fact that \( \angle x = \angle y = \angle z.\)

  The possible answer choices here were:

  i.) Make \(\angle EAC\) not a straight line, but bent at \(A\)
  ii.) Move \(B\) (and its reflection point) closer to \(D,\) keeping \(\overline{EAC}\) straight
  iii.) Make \(\angle x\) a right angle
  iv.) Tilt the whole diagram by \(30^{\circ}\)

  Which of these choices has an effect on whether or not we still \( \angle x = \angle y = \angle z?\)

  For choice ii.) if we move \(B\) closer to \(D,\) keeping the line \(\overline{EC}\) straight, we will still have that \( \angle x = \angle y = \angle z.\) The measure of each angle gets smaller, but they are still all equal to each other.

   
  M2W4D14-ch-part-2-equal-angles-solution3.png
   

  For choice iii.), if we make \(\angle x\) a right angle, then \(\overline{AB}\) will be perpendicular (\(\perp\)) to line \(\overline{AD}.\) Point \(E\) would coincide at the same place as point \(B.\) Then, every angle will be a right angle, so \(\angle x = \angle y = \angle z = 90^{\circ}.\)

   
  M2W4D14-ch-part-2-equal-angles4-solution.png
   

  For choice iv.), if we tilt the whole diagram, this shouldn't change the relationship of the angles. All the angles would be exactly the same size.

  &nbsp
  M2W4D14-ch-part-2-equal-angles5-solution.png
   
   

  If we follow choice i.) and make \(\angle EAC\) a bent line at point \(A,\) then we get a diagram that looks like the one below. As you can see, we have found a change to the diagram that makes \(\angle y \neq \angle z!\)

   
  temp-forum-answer-m2d14.png

  By making line \(EC\) no longer a straight line, we have removed the "X" between lines \(D\) and \(\overline{EC},\) which is the reason for \(\angle y = \angle z\) from before.

  In particular, the "X" formed by the two \(\textcolor{red}{\text{red}}\) lines is what makes it true that \(\angle y = \angle z.\)

   
  M2W4D14-ch-part-2-equal-angles6-solution.png
   

  🙂

• Module 2 Day 14 Challenge Part 3

  Similar Triangles

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  @aaron-wang Thank you so much! 🙂 I've just fixed this.

• Module 2 Day 14 Challenge Part 4

  Shortcut Solution

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  @bravekiwi Yes, this is a really clever trick! You can think of it by first drawing three triangles based on the original path through the rectangle.

  M2W4D14ch4-triple-reflect.png

  Then we can reflect the green triangle and shift it down by one, and shift the purple triangle down twice so that they span a grid that is three times as tall:

  M2W4D14ch4-triple-reflect2.png

  The diagonal line here isn't necessarily a straight one; it happens to look like that because the green line looks like it's almost symmetric. In general, only the shortest path would give a straight line, because that would minimize the distance from start to finish. 🙂

• Module 2 Day 14 Your Turn Part 1

  Shortest Path on a 3-D Cube

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• Module 2 Day 14 Your Turn Part 2

  Unfolding the Cube

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  @victorioussheep Yeah there are a ton of ways to unfold a cube! You can try cutting out a cube net with paper and experiment with it on your own if you want 🙂