# Getting Stuck on a Geometry Problem

I find it helpful (and fun!) to try to understand aspects of problem solving from trying to solve problems on my own. This is a problem that I got stuck on. I had a few ideas to start and then I got stuck.

At the urging of some folks on twitter I used Geogebra to figure out what the value of x is, and now I know and am trying to find a proof. Then, I’ll try to figure out what I would have needed to try to discover the answer on my own, in the first place.

I’ll check back here when I’m ready for spoilers. In the meantime, feel free to use the comments to discuss any aspect of the problem. Here are some prompts:

• What solutions did you find?
• What possible hints could you offer me (or anyone) who gets stuck on this problem?
• What math could one/I learn after working on this problem?
• What strategies/heuristics could come in to play while working on this problem?

Looking forward to reading more when I give up or figure this out!

(Also, this is my first post in this space. Maybe I’ll keep on going, maybe I won’t. Subscribe at your own risk.)

(PS Oy, wordpress sticks ads at the end of posts these days? Sheesh. Maybe I’ll jump ship back to blogger. Or get my own domain. Or something.)

## 13 thoughts on “Getting Stuck on a Geometry Problem”

1. I’ll summarize my interaction with this problem.

My first reaction was that there was not enough information. It seemed like any value for x < 45° would work. To be honest, I persisted in this erroneous belief for quite a while. I said to myself that the upper triangle was irrelevant and irritating.

Michael helped me see my error by out-and-out telling me that what we had was an SSA situation, with an obtuse A, which does yield a unique solution. I was very pleased to have my error pointed out. A hint purist might suggest a slightly lesser hint: AC's length depends on AD's. But I'm not a hint purist, so I was fine with that.

Armed with this new information, I stared at the problem for quite a while, and concluded that the only way in was trig. However that seemed like it was really no fun, while building this in GeoGebra would at least be enjoyable. But then I didn't even build it from scratch: I used an SSA GeoGebra file I had. (http://www.mathedpage.org/constructions/triangles/ssa.html)

That revealed that the answer was 15°. Far from feeling like I was cheating, I felt like I got the necessary boost in confidence. If the answer had turned out to be some weird angle, I would have concluded it was a trig problem, and moved on with my life.

Mysteriously, just knowing this and doing a little more staring, suggested that I should extend CD (on D's side) and drop a perpendicular AE onto it. And there you have it.

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1. ∆ADE is a half-square, with leg 1/sqrt(2), or sqrt(2)/2. ∆ACE is a right triangle with leg AE = half the hypotenuse AC. So ∠CAE = 60°, and x = 15°.

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2. First, I found angle (< from here on out) BAC = < BCA = 45 and that < ADC = 135. I then drew a circle through points ADC, call it c. I extended CB until it hit circle c at E. I then drew segment EA. <AEC is subtended by a 90 deg arc (because the remaining 270 deg arc subtends the 135 angle), so <AEC = 45. Additionally, <ABE = 180 – <ABC = 90, making triangle ABE a right triangle congruent to triangle ABC. This means <EAC is also 90, which implies that EC is a diameter of c and that B is the center of c. But D also lies on circle C, so BD is a radius and is congruent to BA. Additionally it is given that BA is congruent to AD, so BD = AD = BA and triangle ABD is equilateral. Thus, <BAD is 60 and x = 15.

Not sure the best way to share my augmented drawing. I can tweet it if you want.

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3. One thing super-interesting about this problem to me is that the angles are specifically labeled “x” and “45-x” where it might appear simpler to label the bottom angle “135”. Turns out x and 45-x are pretty useful labels that may point the way to one solution method.

I started via Law of Sines then moved to geometry once I’d found an answer. It’s interesting how in a problem like this, determining the answer gives motivation to continue instead of halting it. I wonder if that would be true for most people, though.

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4. I think it is really interesting how our initial attempts shaped the way we solved the problem. I really wanted a geometric solution, so I didn’t even consider trig as an approach. The first thing I did was to do some angle chasing but that turned up fruitless for me. That did reveal another isosceles triangle, but didn’t help me get the answer.

My next approach was to construct the situation in GeoGebra so I could get a sense of the dynamics of it. I constructed it so that point D was free to move but also constrained on a circle to preserve the length congruence given in the problem. I then moved it until the angle at D was 135. To my surprise I saw that this happened precisely when D was on the circle I had used to draw BA and BC. This revealed that triangle BDC was also isosceles (which also revealed the equilateral triangle I ultimately used for the solution), so I went back to angle chasing trying to prove that the base angles were equal. This again proved fruitless.

At this point, I decided to abandon angle chasing and went back to proving that D lies on the circle with center B and radius BA or BC. I really enjoy problems with cyclic quadrilaterals, so whenever I see a quadrilateral problem like this, I like to imagine inscribing it (or parts of it) in a circle, so this approach was pretty natural too me. From here, the solution was pretty straightforward for me.

Henri’s solution is wonderful, and now that I see it, it just jumps out at me. During my problem solving process, however, I barely considered lengths of sides other than those I knew were congruent. I’m wondering if that is because I didn’t try any trig? The trig approach automatically keys you in on the relative lengths of sides, which might have made this approach more apparent to me. On the other hand, I constructed the problem in GeoGebra which required me to use circles in ways I wouldn’t have if I had just drawn the problem on paper, and this clearly clued me on the important circles in the problem. It’s really interesting how, at least for me, my background and initial approaches shaped my ultimate solution. As I said, I now see Henri’s solution clear as day, but his initial hint of extending CD and drawing an altitude didn’t land on me–perhaps because I wasn’t focused on relative side lengths.

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1. I’m guessing that my awareness of the lengths was largely because of Michael’s SSA hint.

The idea to extend CD came from the fact that ∠DCA needed to be 30°, and since I know more about 30° angles than 15° angles it was worth looking at. Thinking of dropping the perpendicular was probably because of the lack of anything else to do.

Note that this would have yielded nothing if I hadn’t been aware of the fact that 1/sqrt(2) = sqrt(2)/2. So Common Core or not, some understanding of manipulating radicals is indeed important in some situations.

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5. I’m not good at GeoGebra (YET… there’s your growth mindset… that’s part of what next summer’s for, after Desmos), so I just did it on paper. I loved Aran’s solution and was able to follow it just fine, but as I sketched as I went along, my circle “center” was not particularly close to B! I was wondering how the circle seemed like the way to go, but once I read it came from the GeoGebra construction, it all made sense.

Similarly, Henri’s half-square wasn’t at all obvious from my scribbling, though I can follow the argument.

Anyway… I am going to add the Law of Sines thing to the comment in case it doesn’t spring out at some interested reader. As stated above, <D must be 180-x-(45-x)=135°. sin 135°/AC = sin(45-x)/AD by Law of Sines. Since AD=AB=BC=AC/sqrt(2) [Pythagorean Thm on top triangle], sin(45-x)=sin135°/sqrt(2)=.5, so 45-x=30 and x=15°.

What I like about the Law of Sines solution is working out what one can derive from what's already there, then using a known relationship to define the rest. I thought it was clean and relatively elegant. What I didn't like is the lack of intuition as I went along. I knew I had an equation with the solution, but when I put it into the calculator I had no intuition in advance that it was 15°. It would have been more satisfying to know 45-x was 30° from some actual geometrical insight.

I still can't quite follow https://twitter.com/mpershan/status/598164713508339712 , Michael… it certainly seems believable, but I can't figure out the order of the reasoning. If it takes you somewhere interesting, maybe you could have mercy on me and spell it out a bit more?

I think I'll sic my clever 8th-grade-Geometry-SBAC-opt-outers on this on Thursday and see what they come up with.

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6. Oooh, I like the similar triangles!

I notice you took it for granted that rotating the diagram could be helpful; I doubt you would have been as likely to put the altitude where you did if it were in the original orientation. By the time students have solved some high school geometry problems, they often but do not always see rotating the diagram as an obvious aid. I think probably students who have used technology a lot for geometry would be more likely to take that step.

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1. Oh, you’re right!

Actually, there’s more to it still. I didn’t think to draw the altitude until I embedded the entire diagram in a square. It was the square-perspective that got me to rotate the diagram. I’m not sure how to describe the square-move other than, if you’re stuck try to embed the diagram in some larger shape (square, circle, etc) that might give you a change of perspective?

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7. It did work. I shared on Twitter because I didn’t know how to post it here.

The truth is that I only made this today. When solving the problem, I was working from a horribly distorted hand-made sketch, which made it quite a bit harder.

Also shared on Twitter: my regret about not seeing that ∆ABD was equilateral, which would have been more obvious on an accurate drawing, given ∠BAD.

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8. This was a fun problem.
Because of the Triangle Sum Theorem, m<D = 135º.
Because of special right triangles and the Pythagorean Theorem, AC = √2(B).
AD = AB = BC was given, so let the length of segment AD be "y" units.
Then AC = y√2 units.
Using the Law of sines:
(y√2)/[sin(135)] = (y)/[sin(45-x)]
By the Division property of equality,
√2/(√2/2) = 1/[sin(45-x)] which simplifies to
2= 1/(sin(45-x)
results in sin(45-x) = 1/2,
Solution: x = 15º.

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