I saw another interesting geometry problem on Twitter. Its replies have a couple excellent solutions. I will post the link later, giving you a chance to try it first,
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cries
30cm²
your complete solution?
Find the area of the 3 right angle triangles then deduct that combined value from the total area. Boom the area of the triangle.
Not so quick, point F I assume can move left towards D or right towards C.
The area of ABM is 16cm squared(8*4)/2 , which is a quarter of the area.
edit common sense check putting F midpoint of CD, gives one triangle ADF with dimensions 8 by 4 giving 16cm squared and another FMC with dimension 4 by 4 giving and area of 8cm squared.
Second edit, common sense 2 Areas of DFA and FCM vary as point F moves, But point F is perpendicular to AM so need to find exact point F
edit my solution.
last edit for silly addition mistake
30 cm squared for the triangle
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My first thought was “oh FFS”…but this one is pretty easy I think. I haven’t really looked at @Mick’s answer to avoid spoilers, but this is how I did it without using trigonometry.
First, let x denote each of the four sides of the square (i.e., AB, BC, AD, and CD), such that BM and CM are both 0.5x because M is the midpoint. Obviously, x = 8 cm because x2 = 64 cm2.
Let f denote the distance DF, such that the distance CF is x−f. Let 1, 2, and 3 denote the outer three right-angled triangles (i.e., ABM, ADF, and CFM, respectively).
Pythagoras and area = (base × height)/2 give the following for the three outer triangles:
Triangle 1:
x2 + 0.25x2 = AM 2
Area = (x × 0.5x)/2 = (8 × 4)/2 = 16 cm2 (this is obvious anyway, because it’s half of the rectangle formed from half of the total square, so 64/4).
Triangles 2 and 3:
AD 2 + DF 2 = AF 2
CF 2 + CM 2 = FM 2
AF and FM are identical, because d is the perpendicular bisector. Therefore,
AD 2 + DF 2 = CF 2 + CM 2
Substituting this equation with stuff we already know gives
x2 + f 2 = (x−f)2 + (0.5x)2
Expanding:
x2 + f 2 = x2 − 2fx + f 2 + 0.25x2
Canceling/rearranging:
0.25x2 = 2fx
0.25x = 2f
f = 0.125x
That is, f = DF = 1 cm and CF = 7 cm.
The area of triangle 2 (ADF) is then
(AD × DF)/2 = (x × 0.125x)/2 = (8 × 1)/2 = 4 cm2
And the area of triangle 3 (CFM) is
(CM × CF)/2 = (0.5x × 0.875x)/2 = (4 × 7)/2 = 14 cm2
Thus, the area of the inner triangle AMF is 64 − (16 + 4 + 14) = 30 cm2, which matches what @Mick got. 
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Don’t have it yet, but I have a plan for solving it.
me too!
My plan is to wait for a consensus answer, and then edit the crap out of all your posts and claim that as my own answer…
actually, i thought from just looking at it that the answer was 1/2 of the original area. M is the midpoint of BC, so if you float F all the way over to the corner D, then it is defining an exact 1/2 as the isosceles triangle ADF. Does the area AFM vary be some sine relationship of angle AFM as F tracks across towards C? I think so. someone else who is closer to school age can work that one out. (or any one of you who has junior high kids!)
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That’s pretty easy to calculate, actually. The areas of ADF and CFM linearly increase and linearly decrease, respectively, as f (= DF) increases from 0 to 8 cm. Specifically, ADF increases from 0 to 32 cm2 and CFM decreases from 16 to 0 cm2 (which makes sense when you think about it, because f = 0 cm makes the triangle ADF have zero area and triangle CFM have one-quarter of the total area, while f = 8 cm makes triangle CFM have zero area and triangle ADF have half of the total area).
The area of ABM remains constant at 16 cm2, which means that the area of AFM (i.e., the answer) decreases from 32 to 16 cm2. The value of f is constrained at 1 cm though by the nature of d.
Using Pythagoras only (no other trigonometry nor angles), we can find the area of the triangle AMF as a function of area of the area of the square ABCD to be
AreaAMF = AreaABCD * 15 / 32 = 64cm² *15/32 = 30cm²
- x the length of a side of the square
- AreaABCD = x²
- E the point where d meets AM
- H the projection of E on the side BC
- G the projection of E on the side CD
- AreaAMF = length(AM) * length(d) / 2
By Pythagoras we calculate length(AM) = √(x² + x²/2) = √(5/4 * x²) = √5/2 * x
Then since we know that E is at the middle point of AM (since d is a bisector), and we know that M is in the middle of CB: then we also know that H is in the middle of MB, and G is in the middle of DC.
This means that the length(EG) = 3/4 x
But also, we can see that triangles AMB and EGF are similar (same angles)! Where AB is similar to EG, and d to AM.
Now since we calculated the formula earlier that length(AM) = √5/2 * length(AB)
(since x =length(AB)),
we can imply that length(d) = √5/2 * length(EG) = √5/2 * 3/4 x = 3√5/8 x
Then we just resolve:
AreaAMF = length(AM) * length(d) / 2
= √5/2 x * 3√5/8 x / 2
= 15/32 x²
= 15/32 AreaABCD
I’ll just have the popcorn.
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I bought too much. You can have some of mine if you want.
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Seems you’re missing 1/32 x², but I can’t point where you went wrong…
Don’t waste it, there’s a shortage
A~B * A~D surface area
-A/M/F surface area
= didnt do well at math, but did well and using mathematicians figures
Good job, guys, but I stomped all of you. Doesn’t look like any of you made use of FA=FM.
Way to think outside the box, but I didn’t have to draw imaginary triangles.
Way to think outside the box, but I didn’t have to draw imaginary lines.
Good job, but I calculated the answer directly, without having to reduce it from two other triangles.