2014 JBMO Shortlist G1

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Source: 2014 JBMO Shortlist G1

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30629 posts

#1 h Oct 8, 2017, 11:26 AM • 3 Y

Y by ItsBesi, Adventure10, Mango247

Let ${ABC}$ be a triangle with $m\left( \angle B \right)=m\left( \angle C \right)={{40}^{{}^\circ }}$ Line bisector of ${\angle{B}}$ intersects ${AC}$ at point ${D}$ . Prove that $BD+DA=BC$ .

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141 posts

#2 h Jan 16, 2018, 7:15 AM • 2 Y

Y by Adventure10, Mango247

can any one post the official solution ?

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643 posts

#3 h Jan 16, 2018, 7:46 AM • 3 Y

Y by Maths-shippuden, Adventure10, Mango247

Solution. easy! Choose a point $K$ on $\overline{BC}$ such that $BD=BK$ . Note that $ADKB$ is cyclic. Hence $AD=DK$ , moreover, by some angle chasing we have $DK=KC$ . Thus $AD=KC$ . And we are done. $\blacksquare$

This post has been edited 2 times. Last edited by ayan.nmath, Jan 16, 2018, 7:49 AM

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141 posts

#4 h Jan 16, 2018, 8:52 AM • 2 Y

Y by Maths-shippuden, Adventure10

ayan.nmath wrote:

Solution. Choose a point $K$ on $\overline{BC}$ such that $BD=BK$ . Note that $ADKB$ is cyclic. Hence $AD=DK$ , moreover, by some angle chasing we have $DK=KC$ . Thus $AD=KC$ . And we are done. $\blacksquare$

nice solution

This post has been edited 1 time. Last edited by GRCMIRACLES, Jan 16, 2018, 8:52 AM

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348 posts

#5 h Jun 6, 2018, 6:33 AM • 1 Y

Y by Adventure10

Oops, when I tried to make a synthetic solution I extended it, no wonder it didn't work. Will using Sine Law give 10 points? It would even fit in 1 page.

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308 posts

#6 h Aug 29, 2018, 1:03 AM • 3 Y

Y by KhayalAliyev, Adventure10, Mango247

Note that $\angle A=100^{\circ}$ . Let $D'$ be a point on $BC$ such that $BD=BD'$ . Hence, $\angle BDD'=\angle BD'D=80^{\circ}$ . Also, $\angle DD'C=100^{\circ}$ and $\angle D'DC=40^{\circ}$ so $DD'=D'C$ . It suffices to prove that $AD=DD'$ . Note that $\angle BAD+\angle DD'B=100^{\circ}+80^{\circ}=180^{\circ}$ so $ADD'B$ is cyclic so $\angle D'AD=\angle DBD'=20^{\circ}=\angle ABD=\angle ADD'$ so $AD=DD'$ , as desired. $\blacksquare$

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974 posts

#7 h Sep 2, 2018, 8:08 PM • 4 Y

Y by Durjoy1729, mufree, Adventure10, Mango247

see figure:

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2125 posts

#10 h Dec 12, 2018, 9:42 AM • 2 Y

Y by Adventure10, Mango247

I am very bad at making synthetic observations, so here's my try at Trig-bashing
Let $BD$ be produced to $A'$ , such, $AD=A'D$ and Let $AB=b=AC$ and $BC=a$
Therefore, $BD=\frac{2ab}{a+b} \cos 20^{\circ}$ and using angle bisector theorem, easy to find out, $A'D=\frac{b^2}{a+b}$
Let's assume $BD+AD=BC \implies \frac{2ab}{a+b} \cos 20^{\circ}+\frac{b^2}{a+b} =a \implies 2ab\cos 20^{\circ} +b^2=a^2+ab$
Now sine rule gives us, $\boxed{a=2b\cos 40^{\circ}} \implies 4b^2\cos 20^{\circ} \cos 40^{\circ}+b^2 =4b^2 \cos ^2 40^{\circ} +2b^2 \cos 40^{\circ}$
Hence, $4\cos 20^{\circ} \cos 40^{\circ}+\cos 60^{\circ}=2\cos ^2 40^{\circ}+\cos 40^{\circ} \implies 2\cos 40^{\circ} \cdot 2\sin 10^{\circ} \sin 30^{\circ}=2\sin 10^{\circ} \sin 50^{\circ} \implies \sin 30^{\circ}=\frac{1}{2} \text{, which is obviously true!!}$ Hence, $\boxed{BD+AD=BC}$

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30629 posts

#11 h Jul 31, 2019, 3:13 PM • 1 Y

Y by Adventure10

my solution posted here before, without words (no cyclic needed)

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209 posts

#12 h Feb 24, 2020, 3:18 PM • 1 Y

Y by Mango247

Use sine rule extensively.

We write $AD=\frac{AB\sin20}{\sin60}, BD=\frac{AB\sin100}{\sin40}$ and $BC=\frac{AB\sin100}{\sin40}$ . Thus, we obtain a trigonometric equation which is easy to solve if we apply the identity of $\sin A+\sin B=2 \sin {\frac{A+B}{2}} \cos {\frac{A-B}{2}}.$
$

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134 posts

#13 h Mar 14, 2021, 2:20 PM

Y by

Does anybody knows who proposed this problem?

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134 posts

#14 h Mar 14, 2021, 2:31 PM

Y by

Also, it seemed me too easy for a sort list proplem...
(Sorry for double posting but I can't edit from my tablet...)

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500 posts

#15 h Aug 24, 2021, 1:38 AM

Y by

$[asy] size(13cm); import olympiad; pair B = (0, 0); pair C = (5, 0); pair A = (2.5, 2); draw(A--B--C--A); label("$A$", A, N); label("$B$", B, W); label("$C$", C, E); pair E = bisectorpoint(A, B, C); pair D = extension(B, E, A, C); draw(B--D); label("$D$", D, NE); [/asy]$

Let's do some trig! But lets do some very easy angle chasing before we do that. We have that $\angle{ABD}=\angle{DBC}=20$ and we know that $\angle{ACB}=40$ so $\angle{BDC}=120$ . Then we know that $\angle{ADB}=60, \angle{BAC}=100$ .

Now notice by LoS we have
$\begin{align*} BD&=\frac{AB \sin(100)}{\sin(60)} \\ DA&=\frac{AB \sin(20)}{\sin(60)} \\ BC&=\frac{AB \sin(100)}{\sin(40)} \end{align*}$
Now let's try to compute $BD+DA=\frac{AB(\sin(100)+\sin(20))}{\sin(60)}.$ From sum to product identities we have $\sin(100)+\sin(20)=2\sin(60)\cos(40)$ . So then $BD+DA=\frac{AB(\sin(100)+\sin(20))}{\sin(60)}=2\cos(40)=2\sin(50)$ Notice the $2$ and the $\sin(\theta)$ which makes us think of the sine double angle identity. So we get $\sin(100)=\sin(50+50)=2\sin(50)\cos(50)=2\sin(50)\sin(40)$ Now we see that $BD+DA=2\sin(50)=\frac{\sin(100)}{\sin(40)}=BC$ $\blacksquare$

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2 posts

#16 h Dec 14, 2021, 5:11 AM

Y by

Consider a point F on BC such that CF=AD .by angle bisector theorem we know AB/BC=AD/DC replacing lengths we know AC/BC=CF/DC or AC/CF=BC/DC hence DFC and ABC are similar and we get our result

This post has been edited 1 time. Last edited by AlirezaSh, Dec 14, 2021, 5:12 AM
Reason: Typoo

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225 posts

#17 h Dec 14, 2021, 6:20 AM

Y by

Select a point $E$ on $BC$ such that $BD=BE$ . Also $AB = AC$ . $ABED$ comes out to be cyclic, hence $\angle DAE= \angle DBE= 20^ \circ$ . similarly $\angle DEA = \angle ABD = 20^\circ$ . So $AD = DE$ . $\angle CDE= \angle DAE + \angle DEA =40^ \circ$ , which follows $\angle DCE = \angle CDE$ hence $DE=CE$ and thus $AD= EC$ . So $BC = BE + EC = BD + AD$ $\blacksquare$ :clap:

:clap:

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974 posts

#18 h Jul 4, 2022, 6:45 AM

Y by

βλεπε σχήμα

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50 posts

#19 h Jul 4, 2022, 8:03 AM

Y by

Let $X$ on $BC$ be a point such that $BD = BX$ , let $Y$ on $BC$ be a point such that $\angle XDY = 20^\circ$ , and let $Z$ on $BD$ be a point such that $DA = DZ$ . It suffices to show $XC = AD$ .

$\angle A = 100^\circ \implies D = 60^\circ \implies \triangle ADZ$ is equilateral so $AZ = AD$ .

$\angle DXB = \frac12 (180 - 20) = 80 \implies \angle DYX = 180^\circ - 20^\circ - 80^\circ = 80^\circ \implies DX = DY$ .

$\angle BDX = \angle DXB = 80^\circ \implies \angle XDC = 180^\circ - 60^\circ - 80^\circ = 40^\circ \implies DX = XC$

$\angle YDC = \angle ZAD = 60 ^\circ \implies AZ \parallel DY$ .

Furthermore, $\angle DYB = \angle DAB = 100^\circ \implies \triangle ABD \cong \triangle YBD$ so $BY = BA$ . Thus $\triangle BAZ \cong \triangle BYZ \implies AD = AZ = ZY$ . Hence $ADZY$ is a rhombus so by combining all the equalities $AD = AZ = DY = DX=XC$ .

This post has been edited 1 time. Last edited by brainee-chan, Jul 4, 2022, 8:03 AM
Reason: formatting

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1162 posts

#20 h Jul 4, 2022, 10:22 AM

Y by

Let $E$ be the point on segment $BC$ such that $BD = BE$ . Then, $\angle BED = \frac{180^{\circ} - \frac{40^{\circ}}{2}}{2} = 80^{\circ} = 180^{\circ} - \angle BAD$ so $ABED$ is cyclic, i.e. $ABC \sim EDC$ . Now, the Angle Bisector Theorem yields $\frac{DA}{DC} = \frac{BA}{BC} = \frac{DE}{DC} = \frac{EC}{DC}$ which implies $DA = EC$ . To finish, we note $BD + DA = BE + EC = BC$ as desired. $\blacksquare$

Remark: The desired length condition intrinsically motivates the construction of $E$ .

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59 posts

DensSv

#22 h Mar 18, 2025, 2:30 PM

Y by

Sol.
Also, the same problem was given to 7th graders in 2010 in Romania, see here https://artofproblemsolving.com/community/c6h492597p2762324

This post has been edited 2 times. Last edited by DensSv, Mar 18, 2025, 2:34 PM

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