## BdMO National 2021 Junior Problem 4

Anindya Biswas
Posts: 188
Joined: Fri Oct 02, 2020 8:51 pm
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### BdMO National 2021 Junior Problem 4

\(ABC\) একটা সূক্ষ্মকোণী ত্রিভুজ। \(P\) আর \(Q\) হলো \(AB\) রেখাংশের ওপর এমন দুটো বিন্দু যেন \(CP \perp AB\) হয় এবং \(CQ\), \(\angle ACB\)-কে সমদ্বিখণ্ডিত করে। যদি \(AC-CB=18\) আর \(AP-PB=12\) হয়, তাহলে \(AQ-QB\) কত?

\$ABC\$ is an acute-angled triangle. Let \$P\$ and \$Q\$ be points on segment \$AB\$ such that \$CP\perp AB\$ and \$CQ\$ bisects \$\angle ACB\$. Given that \$AC-CB=18\$ and \$AP-PB=12\$, Find \$AQ-QB\$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

wa-sickofmath
Posts: 4
Joined: Sat Apr 10, 2021 9:56 pm

### Re: BdMO National 2021 Junior Problem 4

By using Pythagorean theorem,
\$PC^2+PB^2= BC^2\$
And \$PC^2+(PB+8)^2= (BC+12)^2\$
\$or, PC^2+PB^2+16PB+64=BC^2+24BC+144\$
Now when we subtract these two equations we get,
\$PC^2+PB^2+16PB+64-PC^2-PB^2=BC^2+24BC+144-BC^2\$
\$or, 16PB+64=24BC+144\$
\$or, 16PB-24BC=80\$
From this we can say that,
\$16PB>24BC\$
\$or, PB>BC\$
But that can’t happen as \$PB\$ is the perpendicular of the triangle \$APB\$ and \$BC\$ the hypotenuse of it and perpendiculars can never be bigger than hypotenuses.
So, is my logic wrong here? I can't find any flaws. If anybody can, then please point it out. And if there are no flaws then that means that this problem was wrong, right?

Tasin Alam Jon
Posts: 1
Joined: Sat Apr 17, 2021 8:49 pm

### Re: BdMO National 2021 Junior Problem 4

Great.I think you are right.

Marzuq
Posts: 8
Joined: Mon Apr 05, 2021 4:30 pm

### Re: BdMO National 2021 Junior Problem 4

wa-sickofmath wrote:
Thu Apr 15, 2021 11:09 pm
By using Pythagorean theorem,
\$PC^2+PB^2= BC^2\$
And \$PC^2+(PB+8)^2= (BC+12)^2\$
\$or, PC^2+PB^2+16PB+64=BC^2+24BC+144\$
Now when we subtract these two equations we get,
\$PC^2+PB^2+16PB+64-PC^2-PB^2=BC^2+24BC+144-BC^2\$
\$or, 16PB+64=24BC+144\$
\$or, 16PB-24BC=80\$
From this we can say that,
\$16PB>24BC\$
\$or, PB>BC\$
But that can’t happen as \$PB\$ is the perpendicular of the triangle \$APB\$ and \$BC\$ the hypotenuse of it and perpendiculars can never be bigger than hypotenuses.
So, is my logic wrong here? I can't find any flaws. If anybody can, then please point it out. And if there are no flaws then that means that this problem was wrong, right?
I also got this and though the question is wrong. But here is the solution solved by one person in facebook
(just the values are different)
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Solution of National Junior 2021 problem 4
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joydip
Posts: 49
Joined: Tue May 17, 2016 11:52 am

### Re: BdMO National 2021 Junior Problem 4

wa-sickofmath wrote:
Thu Apr 15, 2021 11:09 pm
By using Pythagorean theorem,
\$PC^2+PB^2= BC^2\$
And \$PC^2+(PB+8)^2= (BC+12)^2\$
\$or, PC^2+PB^2+16PB+64=BC^2+24BC+144\$
Now when we subtract these two equations we get,
\$PC^2+PB^2+16PB+64-PC^2-PB^2=BC^2+24BC+144-BC^2\$
\$or, 16PB+64=24BC+144\$
\$or, 16PB-24BC=80\$
From this we can say that,
\$16PB>24BC\$
\$or, PB>BC\$
But that can’t happen as \$PB\$ is the perpendicular of the triangle \$APB\$ and \$BC\$ the hypotenuse of it and perpendiculars can never be bigger than hypotenuses.
So, is my logic wrong here? I can't find any flaws. If anybody can, then please point it out. And if there are no flaws then that means that this problem was wrong, right?

You are completely correct. Those numerical values (\$AC−CB=12\$ and \$AP−PB=8\$) are not possible. So the only way to get any answer to this problem is to solve it without using any unique properties of those two numbers (by treating them as variables). For example, you can consider the solution in the previous post. The error in this question was my fault, and I sincerely apologize for this. Unfortunately, this problem couldn't be dismissed as that would've been more unfair to those who already solved it in this way during the contest.
The first principle is that you must not fool yourself and you are the easiest person to fool.