Length of a c r e a s e crease

Geometry Level 3

A rectangular piece of paper, 24 c m . 24 cm. long by 18 c m 18 cm wide, is folded once in such a way that two diagonally opposite corners coincide. What is the length of the crease(the line made by folding a paper)?


The answer is 22.5.

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Christian Daang by Christian Daang , 20, Philippines

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2 solutions

Christian Daang
Jan 17, 2015

Derivation of the formula to find the crease of a rectangle.

Let the rectangle be A C D C ′ ACDC' . as the figure suggests, the crease is B E BE .

When you fold it, a convex pentagon will be form so,

Let A C = L e n g t h = t AC = Length = t and A C ′ = W i d t h = s AC' = Width = s

As I said a while ago, a convex pentagon is form so, in T r i a n g l e ( A B C C ′ ) Triangle (ABCC') , Let A B = a AB = a and B C C ′ = s 2 + a 2 BCC' = \sqrt{s^{2} + a^{2}}

Since we know that A B + B C C ′ = t AB + BCC' = t , then,

a + s 2 + a 2 = t a+\sqrt{s^{2} + a^{2}} = t

a − t = − s 2 + a 2 a-t = -\sqrt{s^{2} + a^{2}}

a 2 − 2 a t + t 2 = s 2 + a 2 a^{2} - 2at + t^{2} = {s^{2} + a^{2}}

2 a t = t 2 − s 2 2at = t^{2} - s^{2}

a = t 2 − s 2 2 t a = \cfrac{t^{2} - s^{2}}{2t}

Then, B A = t 2 − s 2 2 t BA = \cfrac{t^{2} - s^{2}}{2t} and B C C ′ = t 2 + s 2 2 t BCC' = \cfrac{t^{2} + s^{2}}{2t}

Then, let B E = c r e a s e = F BE = crease = F , A = ∠ E C C ′ D A = \angle ECC'D Since T r i a n g l e ( E D C C ′ ) Triangle (EDCC') is similar to T r i a n g l e ( A B C C ′ ) Triangle (ABCC') , then, it's corresponding parts are s i m i l a r similar so therefore, E C C ′ ECC' is similar to B C C ′ BCC' = t 2 + s 2 2 t \cfrac{t^{2} + s^{2}}{2t}

By using Law of C o s i n e s Cosines in T r i a n g l e ( E B C C ′ ) Triangle (EBCC') .

F 2 = ( t 2 + s 2 2 t ) 2 + ( t 2 + s 2 2 t ) 2 − 2 ( t 2 + s 2 2 t ) 2 ∗ c o s ( 90 − A ) F^{2} = (\cfrac{t^{2} + s^{2}}{2t})^{2} + (\cfrac{t^{2} + s^{2}}{2t})^{2} - 2(\cfrac{t^{2} + s^{2}}{2t})^{2} * cos(90-A)

-> F 2 = 2 ( t 2 + s 2 2 t ) 2 − 2 ( t 2 + s 2 2 t ) 2 ∗ s i n ( A ) F^{2} = 2(\cfrac{t^{2}+s^{2}}{2t})^{2} - 2(\cfrac{t^{2} + s^{2}}{2t})^{2} * sin(A)

-> F 2 = ( t 2 + s 2 ) 2 2 t 2 − ( t 2 + s 2 ) 2 2 t 2 ∗ t 2 − s 2 2 t ∗ 2 t t 2 + s 2 F^{2} = \cfrac{({t^{2} + s^{2}})^{2}}{2t^{2}} - \cfrac{({t^{2} + s^{2}})^{2}}{2t^{2}} * \cfrac{t^{2} - s^{2}}{2t} * \cfrac{2t}{t^{2} + s^{2}}

-> F 2 = ( t 2 + s 2 ) 2 2 t 2 − ( t 2 + s 2 ) 2 2 t 2 ∗ t 2 − s 2 t 2 + s 2 F^{2} = \cfrac{({t^{2} + s^{2}})^{2}}{2t^{2}} - \cfrac{({t^{2} + s^{2}})^{2}}{2t^{2}} * \cfrac{t^{2} - s^{2}}{t^{2} + s^{2}}

-> F 2 = ( t 2 + s 2 ) 2 2 t 2 − ( t 2 + s 2 ) ∗ ( t 2 − s 2 ) 2 t 2 F^{2} = \cfrac{({t^{2} + s^{2}})^{2}}{2t^{2}} - \cfrac{({t^{2} + s^{2})}*({t^{2} - s^{2}})}{2t^{2}}

-> F 2 = ( t 2 + s 2 ) 2 − ( t 2 + s 2 ) ∗ ( t 2 − s 2 ) 2 t 2 F^{2} = \cfrac{({t^{2} + s^{2}})^{2} - ({t^{2} + s^{2}})*({t^{2} - s^{2}})}{2t^{2}}

-> F 2 = ( t 2 + s 2 ) ∗ 2 s 2 2 t 2 F^{2} = \cfrac{({t^{2} + s^{2}})*2s^{2}}{2t^{2}}

-> F = s t ∗ t 2 + s 2 F = \cfrac{s}{t} * \sqrt{t^{2} + s^{2}}

∴ \therefore

length of crease in this problem = 18 24 ∗ 1 8 2 + 2 4 2 \cfrac{18}{24} * \sqrt{18^{2} + 24^{2}}

= 3 4 ∗ 30 \cfrac{3}{4} * 30

= 22.5 c m \boxed{22.5 cm}

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Lu Chee Ket
Jan 17, 2015

18^2 + x^2 = (24 - x)^2 = 24^2 - 48 x + x^2

=> x = (24^2 - 18^2)/ 48 = (24 - 18)(24 + 18)/ 48 = 5.25

Length of crease = SQRT [18^2 + (24 - 2 x)^2] = SQRT (18^2 +13.5^2) = SQRT (506.25) = 22.5

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