Quadrilateral inside a quadrilateral

Geometry Level 3

A B C D ABCD is a convex quadrilateral such that :

A E A D = B F A B = G C B C = H D C D = 1 3 \frac{AE}{AD}=\frac{BF}{AB}=\frac{GC}{BC}=\frac{HD}{CD}=\frac{1}{3}

If A B C D ABCD area is equal to 1 1 and E F G H EFGH area is equal to a b \frac{a}{b} with co-prime a , b a,b find a + b a+b


The answer is 14.

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4 solutions

Let a r e a ( Δ A C D ) = A area(\Delta ACD) = A , a r e a ( Δ A C B ) = B area(\Delta ACB) = B , a r e a ( Δ A B D ) = C area(\Delta ABD) = C , a r e a ( Δ B C D ) = D area(\Delta BCD) = D

Now, a r e a ( Δ E C D ) a r e a ( Δ A C D ) = E D A D = 2 3 a r e a ( Δ E C D ) = 2 3 a r e a ( Δ A C D ) = 2 A 3 \large\frac{area(\Delta ECD)}{area(\Delta ACD)} = \frac{ED}{AD} = \frac{2}{3} \implies area(\Delta ECD) = \frac{2}{3} area(\Delta ACD) = \frac{2A}{3}

Also, a r e a ( Δ E H D ) a r e a ( Δ E C D ) = D H D C = 1 3 a r e a ( Δ E H D ) = 1 3 a r e a ( Δ E C D ) = 1 3 2 A 3 = 2 A 9 E q . 1 \large\frac{area(\Delta EHD)}{area(\Delta ECD)} = \frac{DH}{DC} = \frac{1}{3} \implies area(\Delta EHD) = \frac{1}{3} area(\Delta ECD) = \frac{1}{3}\cdot\frac{2A}{3} = \frac{2A}{9} \quad \cdots Eq. 1

In the similar manner, a r e a ( Δ B F G ) = 2 B 9 E q . 2 a r e a ( Δ A E F ) = 2 C 9 E q . 3 a r e a ( Δ C G H ) = 2 D 9 E q . 4 area(\Delta BFG) = \frac{2B}{9} \quad \cdots Eq. 2 \\ area(\Delta AEF) = \frac{2C}{9} \quad \cdots Eq. 3 \\ area(\Delta CGH) = \frac{2D}{9} \quad \cdots Eq. 4 \\

Adding all these 4 equations we get,

a r e a ( Δ E H D ) + a r e a ( Δ B F G ) + a r e a ( Δ A E F ) + a r e a ( Δ C G H ) = 2 9 ( A + B ) + 2 9 ( C + D ) a r e a ( A B C D ) a r e a ( E F G H ) = 2 9 1 + 2 9 1 = 4 9 a r e a ( E F G H ) = a r e a ( A B C D ) 4 9 = 1 4 9 = 5 9 area(\Delta EHD) + area(\Delta BFG) + area(\Delta AEF) + area(\Delta CGH) = \large\frac{2}{9}(A + B) + \frac{2}{9}(C+ D) \\ \implies \large area(ABCD) - area(EFGH) = \frac{2}{9}\cdot1 + \frac{2}{9}\cdot1 = \frac{4}{9} \implies area(EFGH) = area(ABCD) - \frac{4}{9} = 1 - \frac{4}{9} = \frac{5}{9}

So, a = 5 a = 5 , b = 9 a + b = 14 b = 9 \implies a + b = \boxed{14}

Vimay MarCisse
Jun 29, 2018

Let be a square of side lenght 1 and area 1. According to the statement, we can draw a new square which has the same length as the hypothenuse of the right triangle of side lenght 1/3 and 2/3.

Thus, this length measures: sqrt ((1/3) ^ 2 + (2/3) ^ 2).

Now, by calculating the area of ​​this new square, we find 5/9. Plus 5 + 9 = 14.

X X
Jun 29, 2018

The area of the outside quadrilateral is 1 2 × A B × A D × sin A + 1 2 × C B × C D × sin C = 1 2 × B A × B C × sin B + 1 2 × D A × D C × sin D = 1 \frac12\times AB\times AD\times\sin A+\frac12\times CB\times CD\times\sin C=\frac12\times BA\times BC\times\sin B+\frac12\times DA\times DC\times\sin D=1 A E F = 1 2 × A E × A F × sin A = 1 2 × 1 3 × A D × 2 3 × A B = 2 9 × 1 2 × A D × A B \triangle AEF=\frac12\times AE\times AF\times \sin A =\frac12\times\frac13\times AD\times\frac23\times AB=\frac29\times\frac12\times AD\times AB C G H = 1 2 × C G × C H × sin C = 2 9 × 1 2 × C B × C D × sin C \triangle CGH=\frac12\times CG\times CH\times \sin C=\frac29\times\frac12\times CB\times CD\times\sin C A E F + C G H = 2 9 × ( 1 2 × A B × A D × sin A + 1 2 × C B × C D × sin C ) = 2 9 × 1 \triangle AEF+\triangle CGH=\frac29\times(\frac12\times AB\times AD\times\sin A+\frac12\times CB\times CD\times\sin C)=\frac29\times1 B F G + D E H = 2 9 \triangle BFG+\triangle DEH=\frac29 So the inside quadrilateral is 1 2 9 2 9 = 5 9 1-\dfrac29-\dfrac29=\dfrac59

Relue Tamref
Jun 29, 2018

Instead of finding directly E F G H EFGH area i'll find the triangle areas and subtract from A B C D ABCD

Consider the diagonal B D BD we know that A B D + B C D = 1 ABD + BCD = 1 If you pay attention A B E ABE have the same height as A B D ABD but it's relative base is A D 3 \frac{AD}{3} so its area is equal to A B D 3 \frac{ABD}{3} Now, look at A E F AEF and A E B AEB :

A E F AEF has the same height as A E B AEB but it's base is equal to 2 × A B 3 2 \times \frac{AB}{3} \therefore

A E F AEF area is equal to A E B × 2 3 AEB \times \frac{2}{3} that is equal to A B D × 2 9 {ABD} \times \frac{2}{9} )

If you do the same with triangle B C D BCD you will get the same pattern. The sum of the "opposite" triangles it's equal to A B D × 2 9 + B C D × 2 9 = 2 9 ( A B D + B C D ) = 2 9 ( 1 ) ABD \times \frac{2}{9} + BCD \times \frac{2}{9} = \frac{2}{9}(ABD + BCD) = \frac{2}{9} (1) Skipping unecessary calculations we get that A F E + B F G + C G H + D H E = 4 9 AFE + BFG + CGH + DHE = \frac{4}{9} so E F G H = 1 4 9 = 5 9 EFGH = 1 - \frac{4}{9} = \frac{5}{9}

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