9th grade 2008 Russian Mathematical Olympiad 6th question

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9th grade 2008 Russian Mathematical Olympiad 6th question

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Source: All Russian 2008, Grade 9, Problem 6

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Umut Varolgunes

279 posts

Umut Varolgunes

#1 h May 21, 2008, 9:28 AM • 2 Y

Y by Adventure10, Mango247

The incircle of a triangle $ABC$ touches the side $AB$ and $AC$ at respectively at $X$ and $Y$ . Let $K$ be the midpoint of the arc $\widehat{AB}$ on the circumcircle of $ABC$ . Assume that $XY$ bisects the segment $AK$ . What are the possible measures of angle $BAC$ ?

This post has been edited 1 time. Last edited by djmathman, Dec 29, 2015, 4:30 AM
Reason: latex stuff

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HTA

#2 h May 21, 2008, 11:34 AM • 2 Y

Y by Adventure10, Mango247

sorry , i can't understand how "XY bisects the segment .. " , maybe I misunderstand somehow ?

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1647 posts

#3 h May 21, 2008, 11:51 AM • 2 Y

Y by Adventure10, Mango247

HTA wrote:

sorry , i can't understand how "XY bisects the segment .. " , maybe I misunderstand somehow ?

How I understood it: Line which goes through X and Y bisects the segment AK. I guess there is a triangle in which this can be achieved, but I couldn't construct such an example...

(The text of the problem is practically the same as the text on imo.org.yu...)

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374 posts

Ahiles

#4 h May 21, 2008, 1:35 PM • 2 Y

Y by Adventure10, Mango247

Where can we find all problems from this year?

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#5 h May 21, 2008, 1:38 PM • 2 Y

Y by Adventure10, Mango247

Here you go, Ahiles...

http://www.imomath.com/othercomp/b348f/RusMO08.pdf

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Umut Varolgunes

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Umut Varolgunes

#6 h May 21, 2008, 2:35 PM • 2 Y

Y by Adventure10, Mango247

yup, imo.org.yu is where i got the question. i wonder if there's anyone knows the question from another source.

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#7 h May 21, 2008, 3:00 PM • 2 Y

Y by Adventure10, Mango247

Here's another source: http://olympiads.mccme.ru/vmo/34/2.pdf

But the text is same!

Quote:

Вписанная окружность касается сторон AB и AC треуголь-
ника ABC в точках X и Y соответственно. Точка K сере-
дина дуги AB описанной окружности треугольника ABC.
Оказалось, что прямая XY делит отрезок AK пополам. Че-
му может быть равен угол BAC?

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Umut Varolgunes

279 posts

Umut Varolgunes

#8 h May 21, 2008, 5:31 PM • 1 Y

Y by Adventure10

Thanks, hsiljak.

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#9 h May 21, 2008, 7:20 PM • 1 Y

Y by Adventure10

XY bisects a segment means that the line XY pass through the half point of the segment.

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Umut Varolgunes

279 posts

Umut Varolgunes

#10 h May 22, 2008, 4:55 PM • 2 Y

Y by Adventure10, Mango247

Sorry, there aren't any typos.

I haven't got an elegant solution but with some calculations i solved the question.
Answer

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¬[ƒ(Gabriel)³²¹º]¼

347 posts

¬[ƒ(Gabriel)³²¹º]¼

#11 h May 22, 2008, 5:23 PM • 3 Y

Y by Adventure10, Mango247, Mango247

Call D the point where XY meets AK. Then we must have:

$2 \overline{AD} = \overline{AK} \ \Longleftrightarrow \ 2 \frac{r \cot {\frac{A}{2}} \cos{\frac{A}{2}}}{\sin{\frac{B}{2}}} = R \sin {\frac{C}{2}} \ \Longleftrightarrow$
$\Longleftrightarrow \ \frac{r}{R} \cos^2{\frac{A}{2}}= \sin{\frac{A}{2}} \sin{\frac{B}{2}} \sin{\frac{C}{2}} \ \Longleftrightarrow$
$\Longleftrightarrow \ \cos^2{\frac{A}{2} = \frac{1}{4} \ \Longrightarrow \frac{A}{2} = \{ \frac{\pi}{3}, \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{5 \pi}{3} \} \ \Longrightarrow \ A = \frac{2 \pi}{3}}$

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#12 h May 22, 2008, 6:24 PM • 2 Y

Y by Adventure10, Mango247

Nice problem !

anonymous1173 wrote:

The incircle of a triangle $ABC$ touches the side $[AB]$ , $[AC]$ at the points $X$ , $Y$ respectively. Let $K$ be the midpoint of the arc $AB$

in the circumcircle of $ABC$ . Assume that $XY$ bisects the segment $AK$ . Ascertain the possible measures of angle $\widehat {BAC}$ .

Proof. Denote the midpoint $T$ of the segment $[AK]$ , the intersections $\begin{array}{c} D\in AB\cap CK\\\ S\in XY\cap CK\end{array}$ . If $T\in XY$ , then prove easily that $a>b$ , $a>c$ .

Show easily that $XD=\frac{(a-b)(p-c)}{a+b}$ . Apply the Menelaus' theorem to the transversals :

$\left\{\begin{array}{cccccc} \overline {SYT}\mathrm {\ in\ }\triangle\ ACK : & \frac {SK}{SC}\cdot\frac {YC}{YA}\cdot\frac {TA}{TK}=1 & \implies & \frac {SK}{p-a}=\frac {SC}{p-c}=\frac {CK}{a-c} & \implies & SC=\frac {p-c}{a-c}\cdot CK\\\\ \overline {SYX}\mathrm {\ in\ }\triangle\ ACD : & \frac {SD}{SC}\cdot\frac {YC}{YA}\cdot\frac {XA}{XD}=1 & \implies & \frac {SD}{a-b}=\frac {SC}{a+b}=\frac {CD}{2b} & \implies & SC=\frac {a+b}{2b}\cdot CD\end{array}\right\|\implies$

$\frac {p-c}{a-c}\cdot CK=\frac {a+b}{2b}\cdot CD\implies\frac {p-c}{a-c}\cdot CK\cdot CD=\frac {a+b}{2b}\cdot CD^2$ . From the well-known relations $\boxed {\begin{array}{c} CK\cdot CD=ab\\\\ CD=\frac {2}{a+b}\cdot\sqrt {abp(p-c)}\end{array}}$

obtain $\frac {p-c}{a-c}\cdot ab=\frac {a+b}{2b}\cdot \frac {4abp(p-c)}{(a+b)^2}$ $\implies$ $b(a+b)=(a-c)(a+b+c)$ $\implies$ $a^2=b^2+c^2+bc$ $\implies$ $m(\widehat {BAC})=120^{\circ}$ .

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#13 h Feb 15, 2009, 6:22 PM • 2 Y

Y by Adventure10, Mango247

Wow, guys, I know the trigonometric solutions are quite short, but this is only an angle-chaser.

I'll just give an outline of the solution (it's all angle chasing).

[geogebra]d985fbfae4112099967f34b883f01a8be04a4de7[/geogebra]

First, let $L$ be the midpoint of arc $AC$ . Then $KL$ is parallel to $XY$ . Moreover, $AX = XE$ , $E$ being the intersection of $KL$ and $AB$ .

Second, $KEIB$ is cyclic, so $\angle IEA = 180^\circ - \angle A$ .

Finally, since $IX$ is both altitude and median of triangle $AIE$ , this triangle is isosceles, so $\angle EAI = \angle IEA \iff \frac{\angle A}2 = 180^\circ - \angle A\iff \angle A = 120^\circ$ .

PS: just an aside: for some reason, the Geogebra command 'incircle' doesn't work in my computer (and a forum search showed this to be the case for some other posters); I redid the picture by constructing the angle bisectors.

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Flakky

#14 h Feb 15, 2009, 9:13 PM • 1 Y

Y by Adventure10

Too much work for nothing!!!
Just observe that $\triangle AXS \sim \triangle CIB$ ( $S=XY \cap AK$ ). Hence $AS=\frac{AX.BC}{CI}$ , but $AS=1/2.KI \Longrightarrow 1/2KI.CI=\frac{AX.BC}{CI}.$ Also we know that $KI.CI=2Rr \Longrightarrow Rr=a(p-a) \Leftrightarrow 1=4cos^2(\frac{A}{2})$ $\Longrightarrow A=120^\circ$

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#15 h Feb 10, 2019, 1:44 PM • 2 Y

Y by Adventure10, Mango247

Another synthetic approach.

Let $L$ be the midpoint of $\widehat{AC}$ and let $I$ be incentre of $\Delta ABC$ .
By Fact 5, we have $IK = AK$ and $IL = AL$ .
So, $KL$ is perpendicular bisector of $AI$ . Also $XY$ is perpendicular to $AI$ .
Thus $KL \parallel XY$ .
Let $AI \cap XY = T$ .
It is not difficult to see that $3AT = IT$ . So, $\boxed{\angle{BAC} = 120^{\circ}}$ $\blacksquare$

This post has been edited 2 times. Last edited by MathInfinite, Feb 10, 2019, 1:45 PM

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