The area of the rectangle ABCD is 10.
Points are chosen on the sides of the rectangle such that BM = MC, AN = NP = PD. Diagonal AC intersects NM at point K and MP at point L.
Find the area of triangle MKL.
AB=a, AD=b, ab=10
S(MNP)=((ab)/3)/2=5/3
AN/MC=2/3=>S(ANK)/S(CMK)=4/9
AP/MC=4/3=>S(ALP)/S(MLC)=16/9
S(MKL)=x, S(ANK)=y, S(PNKL)=z, S(MLC)=u
We have:
x+z=5/3
y/(x+u)=4/9 => 9y=4x+4u
(y+z)/u=16/9 =>16u=9y+9z
But AL/LC=AP/MC=4/3 => AL/AC=4/7 and we have
S(ALP)/S(ACD)=(AL/AC)(AP/AD)=(4/7)(2/3)=8/21 => S(ALP)=y+z=(8/21)S(ACD)=40/21
So u=(9/16)(40/21)=15/14
Now 9y-4x=30/7, 4x+4z=20/3( from x+z=5/3) => 9y+4z=30/7+20/3
4y+4z=160/21 implies now 5y=30/7+20/3-160/21 => y=S(ANK)=6/7+4/3-32/21=(18+28-32)/21=14/21=2/3
Another solution, more simpler than the first:
AB=a, AD=b, ab=10
S(MNP)=((ab)/3)/2=5/3
MK/KN=MC/AN=(b/2)/(b/3)=3/2 => MK/MN=3/5
ML/LP=MC/AP=(b/2)/(2b/3)=3/4 =>ML/MP=3/7
S(MKL)/S(MNP)=( MK/MN)(ML/MP)=9/35 =>S(MKL)=(9/35)(5/3)=3/7
REMARK: Clearly, my first solution contains some computing errors! I will correct soon!
Eventually, in the last line equation
6/7+4/3-32/21=(18+21-32)/21 ... there is a small typo
so that this should be
6/7+4/3-32/21=(18+28-32)/21 ... = 2/3 rather than 1/3
Great job - otherwise.
Steftcho P. Dokov Thanks dear Steftcho! S(ANK)=2/3, and, consequently, S(MKL)=3/7.
Wow, sorry! Instead of pressing Edit, to correct the mistake in my first comment, I pressed Delete! Sorry again!
I recovered!
AB=a, AD=b, ab=10
S(MNP)=((ab)/3)/2=5/3
AN/MC=2/3=>S(ANK)/S(CMK)=4/9
AP/MC=4/3=>S(ALP)/S(MLC)=16/9
S(MKL)=x, S(ANK)=y, S(PNKL)=z, S(MLC)=u
We have:
x+z=5/3
y/(x+u)=4/9 => 9y=4x+4u
(y+z)/u=16/9 =>16u=9y+9z
But AL/LC=AP/MC=4/3 => AL/AC=4/7 and we have
S(ALP)/S(ACD)=(AL/AC)(AP/AD)=(4/7)(2/3)=8/21 => S(ALP)=y+z=(8/21)S(ACD)=40/21
So u=(9/16)(40/21)=15/14
Now 9y-4x=30/7, 4x+4z=20/3( from x+z=5/3) => 9y+4z=30/7+20/3
4y+4z=160/21 implies now 5y=30/7+20/3-160/21 => y=S(ANK)=6/7+4/3-32/21=(18+28-32)/21=14/21=2/3.
z=40/21-2\3=26/21 => x=S(MKL)=5/3-26/21=9/21=3/7
I don't understand what is going on! It reappeared!...and I corrected it there too!
Steftcho P. Dokov Liudmyla Hetmanenko The problem remains the same if we replace the rectangle ABCD with a convex quadrilateral ABCD with AD//BC, but my first solution does not work in this case, because S(ACD) can not be evaluated as (1/2)S(ABCD) .
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