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BdMO National 2013: Secondary 5
Posted: Fri Jan 10, 2014 1:36 am
by BdMO
$ABCD$ is a paddy field of trapezoidal shape. Growth of paddy has been uniform everywhere in the field. Farmers are cutting the paddy and piling it in the nearest edge ($AB$, $BC$, $CD$ or $DA$). What is the portion of the total paddy that is piled up in the side $CD$? It is given that, $\angle DAB=\angle ABC=120^{\circ}$, $\angle BCD=\angle CDA=60^{\circ}$, $AB=BC=50$ units.
Re: BdMO National 2013: Secondary 5
Posted: Sun Jan 12, 2014 7:41 pm
by Fatin Farhan
Joining $$A,C$$
$$AC^2=AB^2+BC^2-2AB.BCcos120= 50^2+ 50^2+50*50=3*50^2$$
$$AC=50\sqrt{3}$$.
$$\angle BAC= \angle ACB=30^\circ, \angle BAC=90^\circ, \angle ACB=30^\circ$$.
$$CD=AC/sin60 = 100$$, $$AD=CD/sin30 = 50$$.
Let E and F be the midpoint of AD and BC.
$$EF= (AB+CD)/2= 75$$.
Let OA be the perpendicular from A to BC. OA intersects EF at G.
$$OA=AD*sin60=25\sqrt{3}$$
$$OG=1/2 * OA=25\sqrt{3}/2$$.
Let P and Q be two points on EF and M and N be two points on BC such that EP=FQ=DM=CN=OG.
Now $$(PMNQ)= 1/2 * (PQ+MN)*OG=1/2 * (100-25\sqrt{3} +75- 25\sqrt{3})*25\sqrt{3}/2 $$
$$=(175-50\sqrt{3})*25\sqrt{3}/4 $$
$$(ABCD)= 1/2 * (AB+CD)* OA = 1/2 * (100+50)* 25\sqrt{3}$$
$$= 150*25\sqrt{3}/2$$.
$$(PMNQ):(ABCD)= 175-50\sqrt{3}/300$$.
Re: BdMO National 2013: Secondary 5
Posted: Mon Jan 13, 2014 1:48 pm
by sourav das
I got different solution : $\frac{3125 \sqrt{3}}{4}$ square unit... Ratio $\frac{5}{12}$
Re: BdMO National 2013: Secondary 5
Posted: Mon Jan 13, 2014 2:26 pm
by Siam
I also got 5/12
Re: BdMO National 2013: Secondary 5
Posted: Mon Jan 13, 2014 5:01 pm
by Fatin Farhan
There was a mistake
Fatin Farhan wrote:Let P and Q be two points on EF and M and N be two points on BC such that EP=FQ=DM=CN=OG.
Now $$(PMNQ)= 1/2 * (PQ+MN)*OG=1/2 * (100-25\sqrt{3} +75- 25\sqrt{3})*25\sqrt{3}/2 $$
$$=(175-50\sqrt{3})*25\sqrt{3}/4 $$
$$(ABCD)= 1/2 * (AB+CD)* OA = 1/2 * (100+50)* 25\sqrt{3}$$
$$= 150*25\sqrt{3}/2$$.
$$(PMNQ):(ABCD)= 175-50\sqrt{3}/300$$.
Joining $$A,C$$
$$AC^2=AB^2+BC^2-2AB.BCcos120= 50^2+ 50^2+50*50=3*50^2$$
$$AC=50\sqrt{3}$$.
$$\angle BAC= \angle ACB=30^\circ, \angle BAC=90^\circ, \angle ACB=30^\circ$$.
$$CD=AC/sin60 = 100$$, $$AD=CD/sin30 = 50$$.
Let E and F be the midpoint of AD and BC.
$$EF= (AB+CD)/2= 75$$.
Let OA be the perpendicular from A to BC. OA intersects EF at G.
$$OA=AD*sin60=25\sqrt{3}$$
$$OG=1/2 * OA=25\sqrt{3}/2$$.
Let P and Q be two points on EF such that $$EP=FQ= 25/2$$
So, $$PQ= 75- 25/2 -25/2 = 25$$.
$$(PDCQ) = 1/2 * (PQ+CD)* OG$$
$$= 1/2 * (100+25)* \frac{25 \sqrt{3}}{2} = \frac{3125 \sqrt{3}}{4}$$.
$$(ABCD)= \frac{1}{2} * (AB+CD)* OA = \frac{1}{2} * (100+50)* 25\sqrt{3}$$
$$= 1875\sqrt{3}$$.
$$(PDCQ):(ABCD)=\frac{ \frac{3125 \sqrt{3}}{4}}{1875\sqrt{3}} = \frac{5}{12}$$