A beautiful geometry question. What is the ratio of sides?

This seems like it's impossible to solve at first! But it's a really nice question from a mathematical Olympiad qualifying test in America. Thanks to Anay for the suggestion!

2021 AMC 12B problem 17

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[rdand]

This makes perfect sense, once it's explained. Myself, I wasn't able to even begin coming up with an explanation on my own, but I was able to follow along without much difficulty. Thank you for the video!

[plunger]

This is really difficult. Respect to anyone who can solve this.

[thecursedbunnyy]

I paused and did the exact same method as you and got the answer🤩 This was a nice one..!

[Mathematician]

I extended AD and BC to meet at O, now used similarlity. This easily gave me the answer. I give the equations here

Let H be the height of the trapezium, h be the height of the triangle having area equal to 2 and (H-h) be the height of the triangle having area equal to 4, h2 be the height of traingle ODC. Assume AB/CD=k

AB. H + CD. H=28 - equ no1

CD. h=4 - equ now 2

AB. H - AB. h =8 - equ no 3

Area(OAB) / Area(ODC) =AB^2 / CD^2
Thus, {Area(ODC)+14} /Ar(ODC) =k^2 - equ no 4
Now express area of ODC in terms of h, CD and k, you will get area of ODC equals 1/2 × CD × h/k-1

From these equations, one can get

28/(k+1) +4k=20, from here you get k

[fongalex]

It's quite interesting question. All necessary steps are learnt from secondary school

[petersievert]

I did a similar approach, but did the substitutions in a different way. I came up with 2(3+sqrt(2)) / (4-sqrt(2)) and was a bit stunned looking at the solution.
But funny enough, rationaliziing my "monster" by multiplying both upper and lower part with (4+sqrt(2)), it boils down to 2+sqrt(2), anything but obvious.

[ddjango]

Very neat approach.
3 variables so, 3 equations and finally the quadratic. Very good

[ihori]

Let h be the height of the trapezoid, x - the height of APB. The square of ADCB is (DC+AB)/2 *h.
The square of APB is AB*x/2 = 4; The square of DPC is DC*(h-x)/2=2.
Hence three equasions: h= 28/(DC+AB); x = 8/AB; h-x = 4/DC.
--> 4/DC + 8/AB =( 4AB+8DC)/ (DC*AB) -> 1/DC + 2/AB = (AB+2DC/ (DC*AB) - ->
--> 2DC^2 +AB^2 = 4DC*AB; AB/DC = q ->
--> q^2+2 = 4q --> q = 2+sqr(2) or q = 2-sqr(2), but latter contaradicts the initial statement.

[crazyhitman]

This was a great one! I would love to see more challenging problems like this one in the future!

[smchoi]

Let AB/CD = r and CD = d, so AB = rd. Place the trapezium on the Cartesian plane so that A is at (0, 0) while AB lies on the +ve x-axis, so B is at (rd, 0). Let M(u, v) be the common vertex of the 4 △s partitioning the trapezium. If C is at (p, q), D is at (p+d, q).

Area of △ABM = 4
(1/2)(rd)(v) = 4
rdv = 8 ...(i')
dv = 8/r ...(i)

Area of △CDM = 2
(1/2)(d)(q-v) = 2
dq - dv = 4
dq = 4 + dv = 4 + 8/r (by (i)) ...(ii)

By the "Shoelace Formula",
Area of △DAM = 5
| 0 u p 0 |
(1/2) | | = 5
| 0 v q 0 |
qu - pv = 10 ...(iii)

Similarly,
Area of △BCM = 3
| rd p+d u rd |
(1/2) | | = 3
| 0 q v 0 |
rdq + (p+d)v - qu - rdv = 6
rdq + pv + dv - qu - 8 = 6 (by (i'))
r(dq) + dv = 14 + (qu - pv)
r(4 + 8/r) + (8/r) = 14+10 (by (ii), (i), (iii))
r² - 4r + 2 = 0
r = 2+√(2) (as r > 1)

[jackychanmaths]

Just translate P to a slant edge in parallel to the base
then apply ratios of areas
done

[Maon]

This actually required very basic principles and maths to solve

[girishbishwanath]

I solved it very easily in two minutes.
It's a coincidence that I thought the construction in the very first time.

[genius]

Could someone please explain what theorem or property Presh started to invoke around 1:22? It looks like a leap of logic to me, so apparently there is some property that I either do not know or have forgotten.

[au-ttttx]

Our 9th class problem of triangles and quads

[اممدنحمظ]

تمرين جميل جيد. شرح واضح مرتب . الكتابة والرسم جيدة . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين

[pahbody]

Where did you take the values for x and y? They seem to have come out of nowhere, why are they 2? Why is the area of the trapezoid equals to 28 in the equation? I’m very confused

[frozenturtl]

I feel like making the segments into one variable only at the end is harder for most, I found it easier to make the two bases variables first, I don’t know how much this helps others but works for me

[anon]

Nice.
It's cool that the ratio is fixed and the area is fixed, yet you can get different trapeziums.
as long as h = 4(3-sqrt(2)) / [CD]
there'll be a point P where all the criteria are true.

[jimmykitty]

Thanks a Zettabyte Sir ❤
I love these problems a lot ☺
Love from Bangladesh 🇧🇩🇧🇩🌿🌿