Regional Olympiad - FBH 2018 Grade 11 Problem 3

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Regional Olympiad - FBH 2018 Grade 11 Problem 3

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Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2018

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#1 h Sep 18, 2018, 10:20 PM • 2 Y

Y by Adventure10, Mango247

In triangle $ABC$ given is point $P$ such that $\angle ACP = \angle ABP = 10^{\circ}$ , $\angle CAP = 20^{\circ}$ and
$\angle BAP = 30^{\circ}$ . Prove that $AC=BC$

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#2 h Sep 19, 2018, 4:17 AM • 2 Y

Y by Adventure10, Mango247

We can easily prove $AC=BC=BP$ , all with no trig!

Best regards,
sunken rock

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#3 h Sep 19, 2018, 8:33 AM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Solution

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#4 h Apr 1, 2020, 6:17 PM

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Here is quick and nice trig solution.
From sine law in triangle $ACP$ we get: $\frac{b}{\sin150^\circ}=\frac{AP}{\sin10^\circ}$ or $b=\frac{AP\sin150^\circ}{\sin10^\circ}=\frac{AP}{2\sin10^\circ}$
From sine law in triangle $ABP$ we get: $\frac{c}{\sin140^\circ}=\frac{AP}{\sin10^\circ}$ or $c=\frac{AP\sin140^\circ}{\sin10^\circ}$
Now, from cosine law and using that $\cos50^\circ=\sin40^\circ=\sin140^\circ$ , $a^2=b^2+c^2-2bc\cdot \cos \angle BAC$ we will have $a^2-b^2=c^2-2bc\cdot \cos50^\circ=\frac{AP^2\sin^2 140^\circ}{\sin^2 10^\circ}-\frac{2AP^2 \sin140^\circ \cos50^\circ}{2\sin^2 10^\circ}=\frac{AP^2 \sin^2 140^\circ}{\sin^2 10^\circ}-\frac{AP^2 \sin^2 140^\circ}{\sin^2 10^\circ}=0$ $\implies$ $a^2-b^2=0$ , $a^2=b^2$ and finally $a=b$
$Q.E.D.$

This post has been edited 1 time. Last edited by dikhendzab, Apr 4, 2020, 4:33 PM

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#5 h Apr 3, 2020, 7:20 AM • 1 Y

Y by Mango247

Hint 1
Hint 2
Solution

This post has been edited 3 times. Last edited by Andrew.xyz, Jun 25, 2020, 9:30 AM
Reason: No LaTeX

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#6 h Jul 13, 2023, 2:28 PM

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In∆APB, by sine rule
sin 30°/BP= sin 10°/AP
in ∆ APC , by sine rule,
sin 10°/AP=sin 150°/AC
sin 10°/AP=sin 30°/AC
So, AC=BP
Let angle PBC=x°
So, angle PCB=110°-x
In ∆BPC , by sine rule
sin x/PC=sin(110°-x)/BP
∆ APC , by sine rule,
sin 20°/CP=sin 150°/AC
AC/PC=1/2sin 20°
So, sin(110°-x)/sin x=1/2sin20°
Which gives,
sin(40°-x)=0=sin 0°
So, x=40°
So, angle ABC=50°, angle BAC=50°,
SO, AC=BC
# KRISHIJIVI

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