Problem 6:
$E$ is the midpoint of side $BC$ of rectangle $ABCD$. $A$ point $X$ is chosen on $BE$. $DX$ meets extended $AB$ at $P$. Find the position of $X$ so that the sum of the areas of $\triangle BPX$ and $\triangle DXC$ is maximum with proof.
BdMO National Junior 2011/6 (Secondary 4, H. Sec 3)
BdMO National Junior 2011/6 (Secondary 4, H. Sec 3)
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 Posts: 78
 Joined: Thu Jan 20, 2011 10:46 am
Re: BdMO National Junior 2011/6 (Secondary 4, H. Sec 3)
i use(ABC) for expressing the area of triangle ABC
When X is on the point E (DXC)+(BXP)=(DBC)
In all position of X (DXC)+(BXD)=(BDC)
Now if X is between BE then the extention of DX intersect BC in Q we must say that BQ<BP so always (BXD)>(BXQ)
so if X is between the point B&E then (DXC)+(BXP)<(BDC) and when X is on B then again the sum will same as (BDC) .SO we say that the maximum area is (BDC) only for the point B&E.
The problem has to many solution...........
When X is on the point E (DXC)+(BXP)=(DBC)
In all position of X (DXC)+(BXD)=(BDC)
Now if X is between BE then the extention of DX intersect BC in Q we must say that BQ<BP so always (BXD)>(BXQ)
so if X is between the point B&E then (DXC)+(BXP)<(BDC) and when X is on B then again the sum will same as (BDC) .SO we say that the maximum area is (BDC) only for the point B&E.
The problem has to many solution...........
 FahimFerdous
 Posts: 176
 Joined: Thu Dec 09, 2010 12:50 am
 Location: Mymensingh, Bangladesh
Re: BdMO National Junior 2011/6 (Secondary 4, H. Sec 3)
I solved in the same way. And yes, there's more solution's to it. But I think it's the easiest.
Your hot head might dominate your good heart!
Re: BdMO National Junior 2011/6 (Secondary 4, H. Sec 3)
3. (BPX) + (DXC) = maximum
Through that we get, ½(BP.BX+XC.DC)= maximum
We can see that, BP , XC and BX are interrelated. If, X changes all of these 3 changes.
If, X displaces from E then the area shrinks. But when X is at B and E the area is maximum again. (I cant prove this problem mathematically).
Through that we get, ½(BP.BX+XC.DC)= maximum
We can see that, BP , XC and BX are interrelated. If, X changes all of these 3 changes.
If, X displaces from E then the area shrinks. But when X is at B and E the area is maximum again. (I cant prove this problem mathematically).
r@k€€//

 Posts: 11
 Joined: Tue Dec 14, 2010 1:09 pm
Re: BdMO National Junior 2011/6 (Secondary 4, H. Sec 3)
Let BC=a
CD=b
and BX=l
then if BX=l=a/(2+sqrt(2)) then the sum=ab/2*( (12/(2+sqrt(2))+2/((2+sqrt(2))^2)) / (11/(2+sqrt(2))) ) which is greater then the area of BCD(which is ab/2). I don't know if its the maximum. I found this result using calculus maxima. But looking at the computation and result I don't felling like its a soln of MO. May be I am wrong?
CD=b
and BX=l
then if BX=l=a/(2+sqrt(2)) then the sum=ab/2*( (12/(2+sqrt(2))+2/((2+sqrt(2))^2)) / (11/(2+sqrt(2))) ) which is greater then the area of BCD(which is ab/2). I don't know if its the maximum. I found this result using calculus maxima. But looking at the computation and result I don't felling like its a soln of MO. May be I am wrong?
Re: BdMO National Junior 2011/6 (Secondary 4, H. Sec 3)
In rectangle ABCD, let AB = a and BC = b, BE = 1/2 b
Let X and E be concurrent, then BP will subtend a length of a externally on line AB
As X moves towards B by k units, P moves towards B by ka/b units
Then,
\[Area A = \frac{1}{2} (DC)(CX) + \frac{1}{2} (BP)(BX)
= \frac{1}{2} a (\frac{1}{2}b + k) \frac{1}{2}(\frac{1}{2}b k)(a\frac{ak}{b})=\frac{1}{4}ab + \frac{1}{2}ak + \frac{1}{4}ab  \frac{1}{4}ak + \frac{1}{2}ak + \frac{1}{2}\frac{ak^{2}}{b}
\]
\[Area A = \frac{1}{2} (DC)(CX) + \frac{1}{2} (BP)(BX)
= \frac{1}{2} a (\frac{1}{2}b + k) \frac{1}{2}(\frac{1}{2}b k)(a\frac{ak}{b})=\frac{1}{4}ab + \frac{1}{2}ak + \frac{1}{4}ab  \frac{1}{4}ak + \frac{1}{2}ak + \frac{1}{2}\frac{ak^{2}}{b}
\]
Now differentiating with respect to k,
\[\frac{\mathrm{d} A}{\mathrm{d} k} = \frac{1}{2}a  \frac{1}{4}a  \frac{1}{2}a + \frac{ak}{b}
\]
For max,
\[\frac{\mathrm{d} A}{\mathrm{d} k} = 0,\therefore k =\frac{b}{4} \]
So x is located in the in the midpoint of BE for max area
Let X and E be concurrent, then BP will subtend a length of a externally on line AB
As X moves towards B by k units, P moves towards B by ka/b units
Then,
\[Area A = \frac{1}{2} (DC)(CX) + \frac{1}{2} (BP)(BX)
= \frac{1}{2} a (\frac{1}{2}b + k) \frac{1}{2}(\frac{1}{2}b k)(a\frac{ak}{b})=\frac{1}{4}ab + \frac{1}{2}ak + \frac{1}{4}ab  \frac{1}{4}ak + \frac{1}{2}ak + \frac{1}{2}\frac{ak^{2}}{b}
\]
\[Area A = \frac{1}{2} (DC)(CX) + \frac{1}{2} (BP)(BX)
= \frac{1}{2} a (\frac{1}{2}b + k) \frac{1}{2}(\frac{1}{2}b k)(a\frac{ak}{b})=\frac{1}{4}ab + \frac{1}{2}ak + \frac{1}{4}ab  \frac{1}{4}ak + \frac{1}{2}ak + \frac{1}{2}\frac{ak^{2}}{b}
\]
Now differentiating with respect to k,
\[\frac{\mathrm{d} A}{\mathrm{d} k} = \frac{1}{2}a  \frac{1}{4}a  \frac{1}{2}a + \frac{ak}{b}
\]
For max,
\[\frac{\mathrm{d} A}{\mathrm{d} k} = 0,\therefore k =\frac{b}{4} \]
So x is located in the in the midpoint of BE for max area