HW4 discussion, MA460 Fall 2009, Prof. Walther

• For problem number one, I see that the sum of the areas of the four small triangles is equal to the area of quadrilateral MNPQ. Is this the result we're looking for? It seems a little too obvious but I can't figure anything else out.
  • I also stated for number one that the area of the little paralllogram is half that of the big parallelogram and that the area of all of the little triangles i equal to the area of the little parallelogram

I think that's what they were looking for- Mark

• I am really bad with sketchpad and having trouble with number 2. Can anyone help me get started? -chris
  • sure. at what point are you getting hung up on? - sue]
  • thanks anyways i got it. For any one need help with it, just calculate the lengths and think of thm 28 (multiply)

• any hints for # 10 - sue

• for # 3, if anyone has gotten it, is it a bunch of summing of areas?
  • it is!
• for #3, what areas might we be summing? I remember with the example, the intersection was inside the triangle, so we could use the area of the triangle... any hints? ~Lauren
  • with p being outside, there are now 2 little triangles outside the bigger eq triangle. sum all of the triangles in one direction than all of them in another direction - if that makes sense. what I did was write down the areas for all of the triangles ad see which ones give me what I needed - Sue

• I'm down to 9 and 10 now. no clue on 9 - Sue

-- Check out the thm in the book about having circles... I forget the number, but they have pretty much the exact proof. It helped me get started. And number 10, I drew an extra line through point M which (Honestly, I don't know if you can do) but M is the midpoint and I made it and thought of similar triangles.

• We don't have to do number 9 because we did the proof in class. -Jennie

• Does anyone have a hint on number 5? I'm not sure if it's something obvious and I'm overlooking it or if there is a little trick. Any help would be greatly appreciated!! ~Janelle

• Janelle - for # 5, think areas of litle triangles and big triangles all adding up. and rearrange the equation to show +'s on both sides. makes more sense to me that way. - Sue

• Thanks Sue. As soon as I wrote that I figured it out:-) I just needed to draw one more line so I could make all the connections with the triangles. ~Janelle

• help # 4. I thought I had it, now I don't! Sue

• It seems to me that for number five there about 5 different cases we have to consider, based on whether or not the perpendiculars from P intersect the vertices of triangle. Or does that not have much bearing on the solution? - Tim
  • I'm not sure if that matters or not. My guess is that it doesn't. The reason I say this is because whatever point you choose as the intersection could always be proven to be or not to be the vertex. However, like I said before, I'm not entirely sure. ~Janelle
  • I did make the perpendiculars intersect with the vertices. Only way I could get it to work out. Not sure if this is the cheater way though - Sue

• Any hints for 10? 4? or 5?
  • There is a hint for 10 when you click on the link below: maybe I'm slow, but I don't follow the second hint. Does it have to do with the sides adding up?

to 10:

• • so for number ten if you draw lines through A and D perpendicular to BE you can use similar triangles and parallelograms--Jrhaynie 13:53, 23 September 2009 (UTC)

• for number 5 the different cases are irrelevant. it does not change the fact of what a, b, and c are relative to the respective triangles. If you connect P with the vertices, there will be triangles constructed giving a, b, and c, cool properties...now look at what the areas of the constructed triangles and how they relate to the area of ABC. that should give you a pretty good clue on how to solve the problem.--Jrhaynie 14:24, 23 September 2009 (UTC)

• Does anyone understand what to do based off the hints he gave us? I'm still confused on how to do #10 ~Kat
  • Did you drop a perpendicular line from a to cf and d to cf?- Dana
  • Even if you do the perpendicular line, you end up with this square in the middle. Now, if I can prove the square is really a square, it works.
  • Probably this isn't the most straightforward way to solve it, but when you stumble on a solution, you just take it right? I dropped perpendiculars from A and D to CF, and then considered some sines. Angles ABE and ACF are clearly equal. What are the sines of those angles? That should give you an interesting equality. The same process can be repeated for another equality, and then with a little algebra you'll have your proof. -David
  • I think what he was getting at was that you need to extend the two sides up to make a big triangle. Then you can use similarities and what his hint was to do the proof. The rectangle way works too, it is just a longer proof. -Jennie

10) The way I went about solving 10 was that I had constructed to lines that I had called AG and DH, and they were both perpendiculars to CF. By construction i set AG=DH. Going back to HW3 Problem 5 we can say that with the two lines we have just created we can say that there are similar triangles with the triangle on the left and on the right. With that we know our corresponding ratios between the triangles and from there using algebra we can set the two equations equal to each other meaning we have just proved AB/AC=DE/DF. -Brian

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