The area is constant?
Quadrilateral A B C D has the following properties:
- Opposite angles B and D are each 9 0 ∘ .
- A B and B C have the same length.
- A D + D C = 1 .
What can be stated about other quadrilaterals that have the same properties as A B C D ?
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3 solutions
You need to correct AD + DC = AD + DC' = 1 to AD + DC = AD + AC' = 1
Great observation!
Note: When using prime notation, ideally the base element should be relevant to the prime element. In this case, since C ′ is obtained by rotation D about B by 9 0 ∘ , a more appropriate name will be D ′ .
Very elegant. I think you need to add that your construction of ABCDs encompasses all constructions that fit the properties. This can be achieved simply by rewording the argument: Given a quadrilateral ABCD that fits the properties, it is always possible to construct a unit square around it (as shown in the drawing).
I didn't understand the part where he puts A D + D C = A D + A C ′ = 1 Can someone explain me?
Triangle ADC and Triangle AC'B' is congruent[ assume that the upper side of the big square where extended BC meets it,is named B' ] This is because angle ADC and angle AC'B' are equal. AC=B'C'{being sides of triangles} angle C'B'A=angle DAC
Thus we can say DC=AC'. Therefoore,AD+DC=AD+AC'
Wow! Extremely nice solution!
The problem does not state that AB & BC have a constant length.
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Which is true. They don't have a constant length.
According to hypotenuse length of AC should change each time we rotate
Why AC' and DC are equal?
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Triangle ADC and triangle with points A & C' are the same. 4 triangles on edges are the same.
Note it is a cyclic quadrilateral. Hence AC Is the diameter. Area(ABC)=1/2*x^2. (x=AB=BC) Area(ADC)=1/2mn.(m=AD,n=DC)(m+n=1) Now, m^2+n^2=AC^2=(√2x)^2 Or, (m+n)^2-2mn=2x^2 Or, 1/4=1/2x^2+1/mn It elicits that total area is 1/4 unit
Great solution! But how do you prove that these are the only quadrilaterals that satisfy the condition?
Quadrilaterals of this form consist of a 45-45-90 triangle glued hypotenuse to hypotenuse to a right triangle with legs a and b. Using the Pythagorean theorem, the area of the 45-45-90 triangle is 4 a 2 + b 2 and the area of the other triangle is 2 a b . The area of the quadrilateral is 4 a 2 + b 2 + 2 a b = 4 ( a + b ) 2 . Because a + b = 1 , the area of the quadrilateral is 4 1 .
Very nice.
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How can the sides of both triangles be equal? Or how can you use the same unknown (a&b) for both triangles?
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The line segment AC is the hypotenuse for both triangles, and is length a 2 + b 2
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@John Ross – AC is also length s(2)^1/2 where s is the side of the 45-45-90 triangle. So a and b (or AD and DC) define s (or AB=BC) and there is a unique quadrilateral for every pair of non-negative and non-0 values of a and b (really AD and DC) that satisfy a+b=1 (or AD+DC=1) and can be sides of a right triangle, and they all have the same area. Very nice indeed.
Let AB be equal to x and let AD be equal to y. Then BC will be equal to x and DC wil be equal to 1-y.
By the pythagorean theorem
A
C
2
=
x
2
+
x
2
=
2
x
2
and
A
C
2
=
y
2
+
(
1
−
y
)
2
=
2
y
2
−
2
y
+
1
Since
A
C
2
=
A
C
2
then
2
x
2
=
2
y
2
−
2
y
+
1
The total area of ABCD is
A
r
e
a
(
A
D
C
)
+
A
r
e
a
(
A
B
C
)
A
r
e
a
(
A
D
C
)
=
2
1
y
(1-y)
A
r
e
a
(
A
B
C
)
=
2
1
∗
x
2
A
r
e
a
(
A
B
C
D
)
=
0
.
5
∗
y
∗
(
1
−
y
)
+
0
.
5
∗
x
2
=
0
.
5
∗
(
y
−
y
2
+
x
2
)
2
x
2
=
2
y
2
−
2
y
+
1
−
2
x
2
=
−
2
y
2
+
2
y
−
1
1
−
2
x
2
=
2
∗
(
y
−
y
2
)
y
−
y
2
=
0
.
5
−
x
2
Now we insert this expression into the expression for the area of ABCD:
A
r
e
a
(
A
B
C
D
)
=
0
.
5
∗
(
0
.
5
−
x
2
+
x
2
)
A
r
e
a
(
A
B
C
D
)
=
0
.
5
∗
0
.
5
=
0
.
2
5
I believe there are only two quadrilaterals that fit the specified parameters (mirror images of each other)
that's assuming that sides AB & BC are fixed.
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There was no statement that all four corners lie in the same plane. Allowing that the corners do not need to be in the same plane, e.i., the quadrilateral may be twisted, what can be stated about other quadrilaterals that have the same properties as ABCD? Are their areas also the same?
Imagine the smaller square as a variable square, which can be rotated about within the larger square and so long as the vertices of the smaller square lie on the edges of the larger square, it can be seen that through the variations produced, there are many possible quadrilaterals fulfilling the properties above that can be formed as a result.
Note that AD + DC = AD + AC' = 1 Also, no matter how you rotate the smaller square, the area of ABCD will always remain a quarter of the larger square, i.e 4 1