equal triangle areas (2003-04 Savin Competition 6-8 p4)

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equal triangle areas (2003-04 Savin Competition 6-8 p4)

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#1 h Jun 10, 2021, 6:15 PM • 1 Y

Y by Mango247

Through the midpoints of the diagonals of the convex quadrilateral $ABCD$ , straight lines are drawn parallel to the diagonals. These lines intersect at point $M$ . Prove that the area of triangle $ABM$ is equal to the area of triangle $CMD$ , and the area of triangle $BMC$ is equal to the area of triangle $AMD$ .

https://cdn.artofproblemsolving.com/attachments/c/3/514227ea5289114ec7abb3660d7ce71af81580.png

This post has been edited 3 times. Last edited by parmenides51, Jun 13, 2021, 5:22 PM
Reason: added ABCD as the name of the quadrilateral

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#2 h Jun 10, 2021, 6:16 PM • 1 Y

Y by Mango247

posted for the image link

Attachments:

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manlio

#3 h Jun 13, 2021, 12:37 PM

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Very nice problem. I cannot solve it

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#4 h Jun 13, 2021, 2:13 PM

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manlio

#5 h Jun 13, 2021, 3:16 PM

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$A(ABD) =A(ABM)+A(AMD)$
$A(ABC)=A(ABM)+A(BMC)$

I don't understand these ?????

ABCD is not cyclic

This post has been edited 1 time. Last edited by manlio, Jun 13, 2021, 3:23 PM
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#6 h Jun 13, 2021, 4:22 PM

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manlio wrote:

$A(ABD) =A(ABM)+A(AMD)$
$A(ABC)=A(ABM)+A(BMC)$

I don't understand these ?????
ABCD is not cyclic

$A(ABD) =A(ABM)+A(AMD)$ means Area of triangle ABD = Area triangle ABM + Area triangle AMD

Question has no restriction on the type of convex quadrilateral. According to question, "the area of triangle $ABM$ is equal to the area of triangle $CMD$ , and the area of triangle $BMC$ is equal to the area of triangle $AMD$ " for any convex quadrilateral. It is safe to assume that cyclic quadrilaterals are included. So nothing prevents me from turning the convex quadrilateral into a cyclic quadrilateral.

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manlio

#7 h Jun 13, 2021, 4:28 PM

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Why

$A(ABD) =A(ABM)+A(AMD)$

I think it is

(ABD) = (ABM) + (AMD) + (BMD)

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#8 h Jun 13, 2021, 4:31 PM • 2 Y

Y by centslordm, Mango247

Are you referring to the attached image and using the same letters in the same order? Because BMD is a line from attached image and not a triangle

This post has been edited 2 times. Last edited by biscuit02, Jun 13, 2021, 4:32 PM

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manlio

#9 h Jun 13, 2021, 6:42 PM • 2 Y

Y by Mango247, Mango247

If diagonals of quadrilateral are perpendiculars it is very easy to prove ; now I am trying to prove it when diagonals are inclined and I have embedded the quadrilateral in a parallelogram but I cannot find the right way to prove it

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#10 h Jun 13, 2021, 7:00 PM

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Let $A(0,0),B(2b,0),C(2c,2a),D(2d,2e)$ .

Then $E(c,a)$ and $F(b+d,e)$ .

Point $M(\ \frac{a(b+c+d)(b-d)-ce(b-c-d)}{a(b-d)+ce}\ ,\ \frac{a^{2}(b-d)-ae(b-c+d)+ce^{2}}{a(b-d)+ce}\ )$ .

Areas of $\triangle ABM$ and $\triangle CDM$ both equal to $b \cdot y_{M}$

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manlio

#11 h Jun 13, 2021, 9:24 PM • 1 Y

Y by Mango247

Look at
https://artofproblemsolving.com/community/c6h2589945_have_you_a_nice_solution_for_4_parallelogram_problem

If you can solve it then also our problem is solved

This post has been edited 1 time. Last edited by manlio, Jun 13, 2021, 9:27 PM
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