BdMO 2020 Regional Higher Secondary P1

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saifmd
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BdMO 2020 Regional Higher Secondary P1

Unread post by saifmd » Sun Mar 21, 2021 1:41 pm

In a triangle ABC, AB=30, BC=50 and AC=60. Length of CF = k × Length of BE. Find k?
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Anindya Biswas
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Re: BdMO 2020 Regional Higher Secondary P1

Unread post by Anindya Biswas » Sun Mar 21, 2021 3:46 pm

We know that area of a triangle= base×height/2
So, $2\times\triangle ABC=CF\cdot AB=BE\cdot AC$
$\therefore \frac{CF}{BE}=\frac{AC}{AB}=2$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Dustan
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Re: BdMO 2020 Regional Higher Secondary P1

Unread post by Dustan » Sun Mar 21, 2021 4:34 pm

I used perpendicular lemma😑 never thought like this. :( sad

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Mehrab4226
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Re: BdMO 2020 Regional Higher Secondary P1

Unread post by Mehrab4226 » Sun Mar 21, 2021 7:05 pm

Dustan wrote:
Sun Mar 21, 2021 4:34 pm
I used perpendicular lemma😑 never thought like this. :( sad
What is the perpendicular lemma?
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

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Anindya Biswas
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Re: BdMO 2020 Regional Higher Secondary P1

Unread post by Anindya Biswas » Sun Mar 21, 2021 7:47 pm

Mehrab4226 wrote:
Sun Mar 21, 2021 7:05 pm
Dustan wrote:
Sun Mar 21, 2021 4:34 pm
I used perpendicular lemma😑 never thought like this. :( sad
What is the perpendicular lemma?
Let $A,B,C,D$ be four points in the plane. $AC\perp BD$ if and only if $AB^2-BC^2+CD^2-DA^2=0$. Very important lemma, whoever reading this, try to prove it.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Mehrab4226
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Re: BdMO 2020 Regional Higher Secondary P1

Unread post by Mehrab4226 » Sun Mar 21, 2021 8:50 pm

Anindya Biswas wrote:
Sun Mar 21, 2021 7:47 pm
Mehrab4226 wrote:
Sun Mar 21, 2021 7:05 pm
Dustan wrote:
Sun Mar 21, 2021 4:34 pm
I used perpendicular lemma😑 never thought like this. :( sad
What is the perpendicular lemma?
Let $A,B,C,D$ be four points in the plane. $AC\perp BD$ if and only if $AB^2-BC^2+CD^2-DA^2=0$. Very important lemma, whoever reading this, try to prove it.
The first part is elementary, as we can use the Pythagorean theorem to do so.
For the 2nd part we have,
$AB^2-BC^2+CD^2-DA^2=0$
Let, $AC$ and $BD$ intersect at $O$. And $\angle AOB=\theta$
$AB^2=AO^2+BO^2 -2 \times AO \times BO \times \cos\theta$
$BC^2=BO^2+CO^2 -2 \times BO \times CO \times cos(180- \theta)$
$\therefore BC^2=BO^2+CO^2 +2 \times BO \times CO \times \cos(\theta)$
And the others have similar pattern.
Now,
$AB^2-BC^2+CD^2-DA^2=-2AO\times BO \times \cos(\theta) - \cdots =0$
Or,$\cos(\theta)(-2AO \times BO -2BO \times CO -2CO\times DO -2 DO \times AO)=0$

Which implies $\cos \theta=0$
$\therefore \theta=90^o$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

saifmd
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Re: BdMO 2020 Regional Higher Secondary P1

Unread post by saifmd » Mon Mar 22, 2021 2:56 am

Anindya Biswas wrote:
Sun Mar 21, 2021 3:46 pm
We know that area of a triangle= base×height/2
So, $2\times\triangle ABC=CF\cdot AB=BE\cdot AC$
$\therefore \frac{CF}{BE}=\frac{AC}{AB}=2$
That's an amazing solution. I also solved the problem but used a long naive area based solution. Thank you very much, learned a new way to think.

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