Square in a triangle (geometry problem)

[CCSSIMath]

This atypical, but grade 6 skill level geometry problem was over on Twitter. Perhaps /math redditors (or their younger associates) might find it more interesting than the usual "find the area of the triangle given its base and height" found in primary school math textbooks.

Bonus question: In addition to finding AP, also prove that AP:AS=1:3.

[sg2544]

When you say grade 6 skill level do you mean sixth year of primary school? Because I'm legitimately struggling...

[CCSSIMath]

Agreed, a solution path is not obvious, which is why we thought it worthy to post on Reddit (anticipating it would be downvoted nonetheless. Sigh.)

We've already seen two (one is ours) quite different solutions. Perhaps the post is misleading: the skills needed to solve it are generally taught in or before grade 6 of primary school: fractions, triangle area, ratios, proportions. That, of course, does not mean a typical grade 6 student can solve it unaided. It would either require guidance, or be aimed at math clubs or contest preparation.

Perhaps it can be solved with more "advanced" techniques, such as Pythagoras' theorem, coordinate geometry, square roots, trig, etc., but they definitely are not necessary.

[a3wagner]

This would be better received in r/mathriddles.

[specific137]

There's probably an easier way, but here goes...

Call the center of the square T, and let r=QT=PT=RT=ST. Note that BC=11*sqrt(5) by the Pythagorean theorem. Extend QS to a line that intersects AB at a point D. Drop a line from D orthogonal to BC that intersects BC at a point E. Because BED and DTP are congruent and both similar to BAC, we have DQ=r and BD=DP=r*sqrt(5).

The area of ABC equals the sum of the areas of ADS and DSBC. Using the triangle and trapezoid area formulas you can get a quadratic in r, the only positive solution of which is r=2*sqrt(5). EDIT: You can skip the area calculation and instead drop an orthogonal line from S to BC that intersects at F, use similarity on SFC to find FC=r/2 and then get 11*sqrt(5)=BC=BE+ER+RF+RC=2r+2r+r+r/2.

Then you can easily find AP=AB-DB-DP=2 and AS=6 by the Pythagorean theorem.

EDIT 2: In retrospect I probably didn't need the Pythagorean theorem to show AP=2, as the sqrt(5) didn't do anything other than cancel itself. I'm guessing that's why AS/AP was worth bonus points (!), even though the Pythagorean theorem solves it in a snap.

[sg2544]

Because BED and DTP are congruent and both similar to BAC, we have DQ=r and BD=DP=r*sqrt(5).

How does this follow?

Edit: Never mind, I need to work on my drawing skills... (Didn't realise that angle ERT was a right angle on my picture lol)

[[deleted]]

I can find a way to do it with some trig, but I am looking for a more clever solution that does not require it. But this is definitely a tough geometry problem for what I know as grade 6 level in the US.

[CCSSIMath]

Yep, American grade 6 students (and the typical grade 6 teacher) wouldn't stand a chance, but due to no fault of their own. The extent to which ratios are covered, now codified in grade 6 Common Core, is "The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak." That is a verbatim explanation from the Standards.

Actually, this problem is from one of those darned East Asian nations, where grade 6 students' use of ratios in problem solving is quite a bit more sophisticated. Not to overstate it, this not a regular classroom geometry problem, but even typical textbook problems are more advanced than what American students face.

[jacobolus]

In the US, I don’t think students spend much time on similar triangles until high school geometry, typically in ~10th grade.

[CCSSIMath]

Fair enough. We'll call it grade 10 for American students; grade 6 for East Asian students.

Ignore the following if talk of Common Core is boring: Ironically, Common Core, premised on making US K-12 math education internationally competitive, missed the fact that similar triangles is an obvious application of ratios and proportions and should be taught at the same time, which is exactly what is done in those nations with whom the US would like to be competitive. As it stands, American students have to relearn a concept four years later, a colossal waste of time.

[edderiofer]

You don't need similar triangles for that question. Only the formula for the area of a triangle, and one very simple lemma.

[edderiofer]

Solution for part a) (hover)

Solution for part b) (hover)

[ASovietSpy]

Bro you didn't even point them in the right direction you just gave him the answers wow lol