I don’t know either.
Yeah that’s fine. It seems relevant and on-topic. It’s not always easy or comfortable for me to reply about it, but I am trying to make some effort.
If you think you understand how the shapes fit together and relate to each other, and how that relates to the algebra, then you could solve it using the Pythagorean theorem next. You can revisit my solution again after you do all of them.
I made a start on trying to figure this out. I labeled some points along the black square to make things easier to talk about.
Any errors in the following?
There are a number of similar triangles in the above diagram. One example is triangle KFC and triangle LDC (which encompasses KFC).
They are both right angle triangles which means they have an angle of 90 degrees. They also share an angle (angle KCF or LCD, respectively). Since they have identical angles for two of their angles, their third angle (angle LDC or KFC, respectively) must likewise be the same. Since they have identical angles, they are similar.
Triangle LDC is the larger of the two triangles. We know from a straightforward application of the the information we are given that line FC (the hypotenuse of triangle KFC) is 10 long and line DC (the hypotenuse of triangle LDC) is 20 long. So triangle LDC is a 2X scaled up version of triangle KFC.
I actually didn’t assume that they were both right angle triangles, I used a different assumption to get to them being similar.
I did it a slightly different way:
\triangle DLC \cong \triangle CKB b/c of rotational symmetry, so \triangle FKC \sim \triangle CKB. (Note: \cong means ‘is congruent’).
Since \triangle FKC \sim \triangle DLC \sim \triangle CKB it must be that
\angle FKC = \angle CKB
But also
\angle FKC + \angle CKB = 180^\circ
\therefore \angle FKC = \angle CKB = 90^\circ
The way you did it sounds like you assumed 90^\circ first then didn’t find a contradiction where you might have otherwise (if the angles weren’t equal). Mb just the way you wrote it tho.
EDIT by Elliot to add spoiler tags.
I think you are skipping a major step here? Are you just assuming that the triangles DLC and FKC are similar?
I didn’t write these bits explicitly, but (I think) they are needed for the full chain of reasoning:
DH \parallel FB \\
\therefore \angle DLC = \angle FKC \;
\text{and} \; \angle LDC = \angle KFC \\
\text{observe:} \; \angle DCL = \angle FCK \\
\therefore \triangle DLC \sim \triangle FKC
I think that’s enough.
Ok, I’m not quite clear on the point of your earlier post then.
Like, Justin was assuming the angles were 90 degrees, and using that to figure out that the 2 triangles (DLC & FKC) where similar.
But the stuff you wrote just started with the assumption that the triangles were similar, and then used that to show that the angles were 90 degrees.
If you just assume that the triangles are similar, then you don’t even need to know that they are 90 degrees though - you can solve the problem without them needing to be right angle triangles.
The point was to share a way to show that FKC and LDC are right triangles, like I was responding to Justin’s assumption. I wasn’t talking about the main problem, though (I didn’t show any triangles were right triangles when solving it).
That said, I started writing out the maths too early, so sorta retconed that goal. I didn’t start with a clear goal. That’s probably why my earlier post isn’t v useful or clear.
Yeah. Also, I checked my original working and it literally starts with \triangle BJE \sim \triangle BKC . I let autopilot do that (it seemed ‘obvious’). In hindsight, that sort of skipping-over-stuff means I’m probs not going to make a useful contribution to the topic (with that mindset/method).
Oh, ok.
I wasn’t clear what exactly you meant when you said:
Like, I wasn’t sure if that meant that you approached the whole problem a different way, or what.
But you just meant that you didn’t start with the assumption that the triangles were 90 degrees, and instead started with something different which could be used to prove/show that the triangles were 90 degrees.
It was confusing to me that your solution seemed to be just assuming the triangles were similar, without even stating that explicitly at any point, when that was the thing at issue in the first place - like, Justin was using the 90 degrees thing to figure out that the triangles were similar. So the work that you showed didn’t even show the part that you did differently - like, that you used a different way to get to the triangles being similar.
(Btw, I also used a different way to get to the triangles being similar - I didn’t start with the 90 degrees assumption, or use it at all in my non-pythagorean solution. But I purposely haven’t said what I used, because I think Justin wanted to solve it himself.)
Yeah. I agree it’s unclear (mb misleading)
This didn’t occur to me – like that I started with the thing Justin was trying to figure out. That makes the earlier post a bit pointless, my second post (where the maths starts with DH \parallel FB) is way more useful.
Interesting. I’m not sure if it’d be best to continue along my current approach or try to figure out how you did it. I seemed to want to continue thinking about it from my current approach some. I wrote a bit more:
Here are some facts we know in light of the above analysis:
2 \overline{FC} = \overline{DC} = 20
2 \overline{KF} = \overline{LD} = \, ?
2 \overline{KC} = \overline{LC} = \, ?
\angle DLC = \angle FKC = 90
\angle LDC = \angle KFC = ?
\angle KCF = \angle LCD = ?
So we know some things but we’re still missing some information. In particular, we don’t know the lengths of the sides other than the hypotenuses or the angles other than the right angles. So now it’s time to think about whether we can figure that out given the information we have.
Just to be clear, I actually think it would be a good idea to try to solve the problem using pythagorean theorem first, since that is something you think you already know how to do.
I think it’s OK to continue with your own line of thinking.
The difference is basically how we decided that the triangles were similar. We both decided they were similar triangles. And then my solution depended on that assumption. So you could figure out more & get to the same solution without deciding the triangles were similar in the same way that I did.
It’s ok to figure something out, see if it seems to work, and then go back and check your original assumptions.
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Also think about what other information you know or could figure out about, besides the things you’ve already identified as missing.
I made a start on this but haven’t figured it out yet.
We know that the sides of the square are 20 long. We also have points along the midpoints of those sides, so \overline{GD}, for example, is 10 long.
\triangle GDC has sides \overline{GD}, \overline{DC} and \overline{GC}. We know from the information we have that \overline{GD} is 10 long and that \overline{DC} is 20 long. By the pythagorean theorem:
10^{2} + 20^{2} = c^{2}
500 = c^{2}
10\sqrt{5} = c
So \overline{GC} = 10\sqrt{5}.
\overline{DH} \, , \overline{AE} and \overline{BF} are also 10\sqrt{5}.
To be more clear - my suggestion isn’t that you actually finish the pythagorean theorem solution, necessarily.
Basically, it’s OK to make false starts, it’s ok to try one path and then try a different path.
Trying to follow one solution all the way to the end before you are even willing to look at, entertain, think about, etc, other solutions isn’t a good way to solve math problems.
The thing I thought you were doing wrong was that you were systematically blocking out certain types of thoughts & solution paths as “too easy”, and that was hurting your ability to brainstorm and actually think about the problem.
So I just think you should at least try all the ideas you have, to see how much you can figure out & how far you can get on them. Don’t just ignore/block some of your ideas because you think they are “wrong” in some way. (Like, too easy, not promising, not rigorous enough, etc.) But that doesn’t mean you have to finish every path you start.
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If it seems easy you should finish.
But if it seems hard or you get stuck, then you don’t have to finish.
Basically the point is just: don’t block things.
Oh, and related to this – I think people often dismiss things as “easy” when they aren’t actually easy. But they just don’t actually try them, complete them, etc, so they never figure out that the thing they are saying is “easy” is not actually easy for them.
