<BAC=?, <BCD=90^o, isosceles, AC = BC, BD = 2AD (2012 Polish JMO R1 p2)

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Apr 2, 2025

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lcm(1,2,3,...,n)

lgx57 2

N 12 minutes ago by aidan0626

Let $M=\operatorname{lcm}(1,2,3,\cdots,n)$ .Estimate the range of $M$ .

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lgx57

Today at 7:41 AM

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12 minutes ago

Random Question

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Can anyone help me prove $\lim_{x\rightarrow\infty}(1+\dfrac{1}{x})^x=\sum_{n=0}^{\infty}\dfrac{1}{n!}$ ?

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ReticulatedPython 0

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Let $A$ and $B$ be points on a plane such that $AB=1.$ Let $P$ any point in the plane such that $\frac{AP^2}{BP^2}+\frac{BP^2}{AP^2}=3.$ Prove that $\frac{3-\sqrt{5}}{2} \le AP \le \frac{3+\sqrt5}{2}.$

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AoPS Volume 2, Problem 262

Shiyul 7

N 2 hours ago by Alex-131

Given that $\color[rgb]{0.35,0.35,0.35}v_1=2$ , $\color[rgb]{0.35,0.35,0.35}v_2=4$ and $\color[rgb]{0.35,0.35,0.35} v_{n+1}=3v_n-v_{n-1}$ , prove that $\color[rgb]{0.35,0.35,0.35}v_n=2F_{2n-1}$ , where the terms $\color[rgb]{0.35,0.35,0.35}F_n$ are the Fibonacci numbers.

Can anyone give me hint on how to solve this (not solve the full problem). I'm not sure how to relate the v series to the Fibonacci sequence.

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Shiyul

Today at 4:22 AM

Alex-131

2 hours ago

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<BAC=?, <BCD=90^o, isosceles, AC = BC, BD = 2AD (2012 Polish JMO R1 p2)

parmenides51 2

N Apr 2, 2021 by BMOBoy

A triangle $ABC$ is given with $AC = BC$ . Point $D$ lies on the side $AB$ , where $BD = 2AD$ and the angle $BCD$ is right. Find the measure of the angle $BAC$ .

2 replies

parmenides51

Apr 2, 2021

BMOBoy

Apr 2, 2021

<BAC=?, <BCD=90^o, isosceles, AC = BC, BD = 2AD (2012 Polish JMO R1 p2)

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parmenides51

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parmenides51

#1 h Apr 2, 2021, 7:30 AM

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A triangle $ABC$ is given with $AC = BC$ . Point $D$ lies on the side $AB$ , where $BD = 2AD$ and the angle $BCD$ is right. Find the measure of the angle $BAC$ .

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thegreatp.d

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thegreatp.d

#2 h Apr 2, 2021, 8:28 AM

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Am I right?

Let $E$ be the mid point of $BD$ and $\angle BAC=\angle ABC=\theta$
$\implies \angle DEC=2\theta$
Now in $\triangle AEC$ and $\triangle BDC$
$\angle EAC=\angle DBC$
$AC=BC,AE=BD$
So, $\triangle AEC\cong \triangle BDC$
So, $CD=CE$
Again, $CE=DE=BE$ as $E$ is the circumcenter of $\triangle BDC$
So, $CD=CE=DE$
$\implies \triangle BDC$ is equilateral. So $\angle DEC=2\theta =60^{\circ}$
$\implies \theta=30^{\circ}$
$\implies \angle BAC=\theta=\boxed{30^{\circ}}$

This post has been edited 4 times. Last edited by thegreatp.d, Apr 2, 2021, 8:33 AM
Reason: Typo

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BMOBoy

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BMOBoy

#3 h Apr 2, 2021, 9:15 AM

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You could drop the perpendicular from C and that creates a some similar triangles. If we call the foot of the perpendicular E, then, with the use of the similar triangle ratios, one yields that DEC is a 30-60-90 triangle. Then it is okay from there.

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