Square ABCD is divided at its midpoints E, F, G, H. The 4 segments connecting each vertex with the midpoint of the clockwise side bound a region IJKL, which is shaded in the diagram below. What fraction of the square is shaded?

As usual, watch the video for a solution.

What Fraction Of The Square Is Shaded?

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To What Fraction Of The Square Is Shaded?

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

We first establish IJKL is a square.

If we rotate the entire diagram by 90 degrees clockwise from its center, then the diagram would remain unchanged. The segments connecting a vertex to the clockwise side’s midpoint are 90 degree rotations of each other, establishing that IJKL has 4 right angles. The sides of IJKL are the respective distances between the segments connecting a vertex to a midpoint, and those distances must be equal so the sides of IJKL are equal. Thus IJKL is a quadrilateral with 4 right angles and 4 equal sides and is a square.

Method 1: similar triangles

Let IL = x.

The right triangle AID is similar to HLD. By similarity, since H is the midpoint on the hypotenuse AD, L must be the midpoint of DI. So DL = IL = x. Triangles DLH and AIE are congruent by symmetry, so AI = x.

We can calculate the area of the square in 2 different ways.

Triangles DIA, AJB, BKC, and CLD are congruent, each with area 0.5(2x)x = x². Square IJKL has an area of x², so the total area of the square ABCD is:

4x² + x²
= 5x²

The fraction of the square shaded is thus:

area(IJKL)/area(ABCD)
= x²/(5x²)
= 1/5

Alternately we could have seen that AID is a right triangle so we have:

AD²
= AI² + DI²
= (2x)² + x²
= 4x² + x²
= 5x²

So once again we have:

area(IJKL)/area(ABCD)
= x²/(5x²)
= 1/5

Method 2: parallelogram area

We will compute the area of the parallelogram DEBG in two different ways.

Let DG = s, so then DH = AH = AE = s. Then we have:

area(DEBG)
= (base DG)(height upon DG)
= (s)(AD)
= (s)(2s)
= 2s²

Now let LK = x. This is a height upon the base DE of the parallelogram. But DE is a hypotenuse of triangle DAE, so we have:

DE²
= DA² + AE²
= (2s)² + s²
= 4s² + s²
= 5s²

Thus we have DE = s√5 and the area of the parallelogram is:

area(DEBG)
= (base DE)(height upon DE)
= (s√5)(LK)
= (s√5)(x)
= xs√5

Setting the two expressions of the area of the parallelogram equal we get:

xs√5 = 2s²
x = 2s/√5

We now find the area of the square IJKL is:

x² = 4s/5

So we have:

area(IJKL)/area(ABCD)
= (4s/5)/((2s)²)
= (4s/5)/(4s)
= 1/5

Method 3: re-arranging areas

There is also an elegant outside the box solution.

We can re-arrange the areas of the small triangles to join with the trapezoids to form a square. (I justify this in detail in the video–but from the above results we would have the new shape has 3 right angles with 3 equal sides, so this new shape would be a square).

The square ABCD has the same area as 5 squares of the size IJKL. Thus it is easy to see the fraction of ABCD that is shaded is 1/5.

References

Reddit AskMath
https://www.reddit.com/r/askmath/comments/1vg9sla/how_to_solve_this/

Variation problem
https://mindyourdecisions.com/blog/2020/05/17/the-square-inside-the-square/

Math StackExchange
https://math.stackexchange.com/questions/2518341/area-of-a-square-inside-a-square-created-by-connecting-point-opposite-midpoint

Thanks to Willem for the suggestion!

Point E is exterior to square ABCD with EA = 6, EB = 3, and EC = 5. What is the area of square ABCD?

As usual, watch the video for a solution.

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Area Of Square From An External Point

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Suppose the square has a side length equal to x.

Suppose the horizontal distance from B to E is a and the vertical distance is b. We can then construct 3 triangles from point E whose hypotenuses are the three lengths 6, 3, and 5.

By the Gougu Theorem, we have the following three equations:

(I)
a² + b² = 3² = 9

(II)
(a + x)² + b² = 5² = 25

(III)
a² + (b + x)² = 6² = 36

In equation (II) we will solve for a. We will use a² + b² = 4 from (I) in the second step.

(a + x)² + b² = 25
x² + 2ax + a² + b² = 25
x² + 2ax + 9 = 25
a = (16 – x²)/(2x)

We will similarly solve for b in (III):

a² + (b + x)² = 36
a² + b² + 2bx + x² = 36
9 + 2bx + x² = 36
b = (27 – x²)/(2x)

We can now substitute for a and b in (I):

a² + b² = 9
((16 – x²)/(2x))² + ((27 – x²)/(2x))² = 9
(16 – x²)² + (27 – x²)² = 36x²
256 – 32x² + (x²)² + 729 – 54x² + (x²)² = 36x²
2(x²)² – 122x² + 985 = 0

The area of the square is x². The last equation is quadratic in x², so we can solve for its value using Brahmagupta’s quadratic formula, giving:

x² = [-(-122) ± √((-122)² – 4(2)(985))]/(2(2))
x² = 61/2 ± (√1751)/2

Usually in geometry problems we want the positive radical to get a positive length. But in this problem we have two positive values:

x² = 61/2 + (√1751)/2 ≈ 51.4
x² = 61/2 – (√1751)/2 ≈ 9.6

The key in this problem is we want E to be exterior to the square, so we want a and b to be non-negative values. Recall that:

a = (16 – x²)/(2x)

For a to be non-negative, we need x² ≤ 16.

Thus we must exclude the area ≈ 51.4 case. The correct answer is thus:

Area = x² = 61/2 – (√1751)/2 ≈ 9.6 units squared.

This problem was on Reddit AskMath and the picture has a credit to Logic Booster. The poster thought there was insufficient information. But people replied that it was possible to solve. Can you figure it out?

As usual, watch the video for a solution.

Right Triangle Area Puzzle

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Right Triangle Area Puzzle

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Method 1: Algebra

Let the legs be a and b and the hypotenuse c = 64 cm. Because the perimeter is 150 cm we have:

a + b + c = 150
a + b + 64 = 150
a + b = 86
(a + b)² = 86²
a² + b² + 2ab = 86²

But since we have a right triangle we have:

a² + b² = c² = 64²

Substituting we have:

64² + 2ab = 86²
2ab = 86² – 64²
ab/2 = (86² – 64²)/4
ab/2 = (86 + 64)(86 – 64)/4
ab/2 = 825

So the area is 825 cm², and we didn’t have to solve for the sides of the triangle!

Method 2: Inradius

An alternate method is to construct the incircle with a radius r. In any triangle the area is:

rp/2

Furthermore, in a right triangle the hypotenuse is equal to:

c = a – r + b – r

So we can add c to both sides and get:

2c = a + b + c – 2r
2c = p – 2r
r = (p – 2c)/2

So now let’s solve the original problem.

We can solve for the inradius using the given information:

r = (p – 2c)/2
r = (150 – 2(64))/2
r = 11

Then we have the area of the triangle is:

rp/2
= 11(150)/2
= 825

So we again get the answer of 825 cm².

Extra credit

Solve for the lengths of the legs.

Reference

Reddit AskMath
https://www.reddit.com/r/askmath/comments/1tu6zwa/found_on_facebook_not_enough_info/