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BdMO 2020 Regional Higher Secondary P1

https://matholympiad.org.bd/forum/viewtopic.php?f=21&t=6177

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BdMO 2020 Regional Higher Secondary P1

Posted: Sun Mar 21, 2021 1:41 pm
by saifmd
In a triangle ABC, AB=30, BC=50 and AC=60. Length of CF = k × Length of BE. Find k?
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Re: BdMO 2020 Regional Higher Secondary P1

Posted: Sun Mar 21, 2021 3:46 pm
by Anindya Biswas
We know that area of a triangle= base×height/2
So, $2\times\triangle ABC=CF\cdot AB=BE\cdot AC$
$\therefore \frac{CF}{BE}=\frac{AC}{AB}=2$

Re: BdMO 2020 Regional Higher Secondary P1

Posted: Sun Mar 21, 2021 4:34 pm
by Dustan
I used perpendicular lemma😑 never thought like this. :( sad

Re: BdMO 2020 Regional Higher Secondary P1

Posted: Sun Mar 21, 2021 7:05 pm
by Mehrab4226
Dustan wrote: ↑
Sun Mar 21, 2021 4:34 pm
 I used perpendicular lemma😑 never thought like this. :( sad
What is the perpendicular lemma?

Re: BdMO 2020 Regional Higher Secondary P1

Posted: Sun Mar 21, 2021 7:47 pm
by Anindya Biswas
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.

Re: BdMO 2020 Regional Higher Secondary P1

Posted: Sun Mar 21, 2021 8:50 pm
by Mehrab4226
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 $\cos \theta=0$
$\therefore \theta=90^o$

Re: BdMO 2020 Regional Higher Secondary P1

Posted: Mon Mar 22, 2021 2:56 am
by saifmd
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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