peruvian anglechasing (2019 Peru L1 p3 ONEM)

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High School Math Grades 9-12, Ages 13-18

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#1 h Mar 1, 2020, 9:47 AM

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Let $ABC$ be a triangle. On the side $AB$ lies point $D$ is chosen and on the side $AC$ lies point $E$ such that $AE = ED = DB$ and $EC = CB =\frac{AB}{2}$ . Find the measure of the angle $\angle BAC$ .

This post has been edited 3 times. Last edited by parmenides51, Sep 2, 2022, 5:50 AM

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#2 h Mar 1, 2020, 4:02 PM

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are you sure the question is correct because i keep getting 0?

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#3 h Mar 1, 2020, 4:31 PM

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Rule of sines in $\triangle ABC\ :\ \frac{AB}{\sin 3\alpha}=\frac{\frac{AB}{2}}{\sin \alpha}$ .

$2\sin \alpha=\sin 3\alpha=3\sin \alpha-4\sin^{3}\alpha$ .
$4\sin^{2}\alpha=1$ and $\alpha=30^{\circ}$ .

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Arksin

#4 h Mar 2, 2020, 3:13 PM

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Walkthrough of vanstraelen's solution:

Let's call $\widehat{BAC}$ $\theta$ .

Using the multiple isosceles triangles and the fact that the sum of the angles of a triangle is equal to $\pi$ rad, we can easily calculate that $\widehat{ACB}$ = $\pi -3\theta$ .

Using the law of sines in $\bigtriangleup ABC$ , with $AB=c$ and $BC=\dfrac{c}{2}$ , we obtain:

$2sin(\theta)=sin(\pi-3\theta)$ Since $sin(\pi-3\theta)=sin(3\theta)$ ,
$2sin(\theta)=sin(3\theta)$
By the triple angle formula, we know that
$sin(3\theta)=3sin(\theta)-4sin^3(\theta)$ Substituting this into the previous equation gives
$2sin(\theta)=3sin(\theta)-4sin^3(\theta)$ $sin(\theta)=4sin^3(\theta)$ Using $x=sin(\theta)$ , we have
$x=4x^3$ $x-4x^3=0$ $x(1-4x^2)=0$
There are now two possibilities:

1. $x=0$ , which is not possible since our angle is not null
2. $1-4x^2=0$ which has two solutions : $\dfrac{1}{2}$ and $-\dfrac{1}{2}$

Therefore, $\theta$ is an angle whose sine equals either $\dfrac{1}{2}$ or $-\dfrac{1}{2}$ , which is an angle of
$\boxed{{\dfrac{\pi}{6}rad} = {30^{\circ}}}$

This post has been edited 3 times. Last edited by Arksin, Mar 2, 2020, 8:40 PM
Reason: formatting

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#5 h Mar 26, 2020, 5:32 PM

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Synthetic: Let $m(\widehat{ABE})=\alpha$ , then $m(\widehat{ADE})=m(\widehat{BAC})=2\alpha$ . Since $BCED$ is a kite, we get $m(\widehat{DBC})=m(\widehat{DEC})=4\alpha$ , that is, $m(\widehat{ABC})=2m(\widehat{BAC})\ (\ 1\ )$ , so if the perpendicular bisector of $AC$ intersects $AB$ at $O$ we have $m(\widehat{BOC})=4\alpha=m(\widehat{ABC})$ , that is, $BC=OC=OA$ , but $BC$ is half of $AB$ , thence $O$ is midpoint of $AB$ , consequently $m(\widehat{ACB})=90^\circ$ . With $(1)$ we easily find $\alpha=15^\circ\implies m(\widehat{BAC})=30^\circ$ .

Best regards,
sunken rock

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