View topic - IMO 2008, Question 1 • Art of Problem Solving

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orl

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Posted: Jul 16, 2008, 6:24 am • # 1

Let $H$ be the orthocenter of an acute-angled triangle $ABC$ . The circle $\Gamma_{A}$ centered at the midpoint of $BC$ and passing through $H$ intersects the sideline $BC$ at points $A_{1}$ and $A_{2}$ . Similarly, define the points $B_{1}$ , $B_{2}$ , $C_{1}$ and $C_{2}$ .

Prove that six points $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ and $C_{2}$ are concyclic.

Author: Andrey Gavrilyuk, Russia

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Last edited by orl on Jul 20, 2008, 1:14 am, edited 4 times in total.

Jan

Posts: 606
Location: Belgium
Belgium

Posted: Jul 16, 2008, 6:47 am • # 2

Call $M_1,M_2,M_3$ the midpoints of $BC,AC$ and $AB$ .

We know that $BP \perp AC$ , so $BP$ and $M_1M_3$ are also perpendicular, and $B$ lies on the radical axis of $\Gamma_A$ and $\Gamma_C$ . It follows that $BC_1\cdot BC_2 = BA_1 \cdot BA_2$ , so $A_1,A_2,C_1$ and $C_2$ all lie on a circle, whose center is clearly $O$ , the circumcenter of $\Delta ABC$ .

Now we can show that $B_1$ and $B_2$ also lie on the circle centered at $O$ , passing through $A_1$ and $A_2$ , from which the conclusion follows.

TomciO

Posts: 528
Location: Poland, Cracow.
Poland

Posted: Jul 16, 2008, 6:52 am • # 3

The radical axis of $\Gamma_{A}$ and $\Gamma_{B}$ is perpendicular to the line connecting the centers of these circles, i.e. it is perpendicular to $AB$ . Since both circles pass trough $H$ their radical axis is a height from $C$ , so $C$ lies on a radical axis. It means that $CA_1CA_2 = CB_1CB_2$ so $A_1, A_2, B_2, B_1$ lie on one circle. From the same reasoning $B_1, B_2, C_2, C_1$ and $C_1, C_2, A_2, A_1$ lie on a one circle as well. Suppose that this three circles doesn't coincide. Then we obtain a contradiction since radical axis of these circles - the sides of the triangle $ABC$ - don't intersect in one point.

Lepuslapis

Posts: 33
Location: Luxembourg

Posted: Jul 16, 2008, 6:59 am • # 4

Another possibility to prove that, let's say, $A_1A_2B_1B_2$ is cyclic consists in transforming the relation $CA_1\cdot CA_2 = CB_1\cdot CB_2$ into $CD^2 - DH^2 = CE^2 - EH^2$ (where $D$ and $E$ are the respective midpoints of $[BC]$ and $[AC]$ ). Adding $CH^2$ on both sides and applying the generalized version of Pythagoras' theorem (the one with cosines, I ignore the correct English name), we get a condition which is obviously true (simplifying cosines in right-angled triangles)...

plane geometry Posts: 471	Posted: Jul 16, 2008, 7:11 am • # 5 A,B,C lie on the radix axis of each two circles so conclusion apparently follows

TTsphn

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Posted: Jul 16, 2008, 7:17 am • # 6

I have a same solution with you .
Call $M,N,P$ is the midpoint of $BC,CA,AB$
From condition we have :
$BA_1.BA_2=BM^2-HM^2$
$BC_1.BC_2=BN^2-HN^2$
But from $MN||AC,BH\perp AC$ therefore $BH\perp MN$
It gives $BM^2-BN^2=HM^2-HN^2$
Therefore $A_1,B_1,A_2,B_2$ are cyclic on circle $O_c$
Similar for $A_1,A_2,C_1,C_2$ lie on $O_b$ and $B_1,B_2,C_1,C_2$ lie on circle $O_a$
Easy to check that three circle are coincide .
So problem claim.

pohoatza

Posts: 1100
Location: Bucharest
Romania

Posted: Jul 16, 2008, 7:41 am • # 7

Indeed, the proof involving the three radical axis is the best. In any case, the circle belongs to Droz-Farny as Darij well mentioned (see http://www.pandd.demon.nl/drozf.htm; scroll down till figure 3).

By the way, it is obvious that the triangle doesn't need to be acute-angled. I guess the coordinators didn't want messy diagrams :roll:

.

In addition, here is a natural generalization:

Theorem. Let $P$ , $Q$ be two isogonal points with respect to a given triangle $ABC$ . Let $P_{A}$ , $P_{B}$ , $P_{C}$ be the orthogonal projections of $P$ on the sidelines $BC$ , $CA$ and $AB$ , respectively. Denote by $\Gamma_{A}$ the circle centered at $P_{A}$ which passes through $Q$ and let $A_{1}$ , $A_{2}$ be the intersections points of $\Gamma_{A}$ with the sideline $BC$ . Similarly, define $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ . Then, the points $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ are on a same circle centered at $P$ .

I guess it would have been a better IMO problem, at least to avoid the "well-known"-type commentaries.

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Ahiles

Posts: 356
Location: Chisinau, Republic of Moldova
Moldova, Republic of

Posted: Jul 16, 2008, 9:12 am • # 8

It is enough to prove that points $A_1,A_2,B_1,B_2$ are concyclic. Let $O$ be circumcenter of the triangle $ABC$ . We'll prove that $O$ is circumcenter of quadrilateral $A_1A_2B_1B_2$ . Then $O$ is situated on the perpendicular bisectors of the segments $A_1A_2$ and $B_1B_2$ $\Longrightarrow OA_1 = OA_2, OB_1 = OB_2$ . So it is enough to prove that $OA_1 = OB_1$ .
Call $r_1,r_2$ circumcenters of the $\Gamma_a$ and $\Gamma_b$ respectively and $A',B'$ midpoints of the segments $BC,CA$ . Then $\left\|\begin{array}{cc} OA_1 = r_1^2 + OA'^2 \\OB_1 = r_2^2 + OB'^2 \end{array} \right\|$ .
By Stewart we have
$\boxed{\begin{array}{cc} 4r_1^2 = 2(CH^2 + BH^2) - a^2 = \\= 8R^2(\cos^2{B} + \cos^2{C}) - 4R^2\sin^2{A} = \\= 4R^2(2\cos^2{B...$ .

By analogy $4r_2^2 = 4R^2(2\cos^2{A} + 2\cos^2{C} - \sin^2{B})$ .

But $\left\| \begin{array}{cc} OA'^2 = \frac {1}{4}a^2\cos^2{A} = R^2\cot^2{A} \\OB'^2 = \frac {1}{4}b^2\cos^2{B} = R^2\cot^2{B} \...$ .

Then $OA_1^2 = 4R^2(2\cos^2{B} + 2\cos^2{C} - \sin^2{A} + \cot^2{A})$ and $OB_1^2 = 4R^2(2\cos^2{A} + 2\cos^2{C} - \sin^2{B} + \cot^2{B})$ .

So we have to prove that:
$\boxed{\begin{array}{cc} 2\cos^2{B} + 2\cos^2{C} - \sin^2{A} + \cot^2{A} = 2\cos^2{A} + 2\cos^2{C} - \sin^2{B} + \cot^2{B} \\...$
and we are done !!!! :!:

cyshine

Posts: 177
Brazil

Posted: Jul 16, 2008, 9:29 am • # 9

Fix $O$ as center of a coordinate vector system, so that $H = A + B + C$ . If $M = \frac{A + B}2$ we have to prove that $MC_1^2 + OM^2$ is symmetric in $A$ , $B$ and $C$ (it would be the square of the radius of the circle).

But

$MC_1^2 + OM^2 = HM^2 + OM^2 = \left(A + B + C - \frac{A+B}2\right)\cdot \left(A + B + C - \frac{A+B}2\right) + \left(\frac{A+...$

and we are done.

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msecco

Posts: 101
Location: Rio de Janeiro
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Posted: Jul 16, 2008, 10:26 am • # 10

No no no!!
More one trivial Geometry.
Denote AB=2c, AC=2b and BC=2a.
Denote BA_1=x and AC_1=y.Denote too M, N and P the midpoints of BC, AB and AC respectively.
Then, MA_1=a-x=MA_2. We have to prove that BA_1.BA_2=BC_2.BC_1.
Applying the Cosines'Law in triangles BNH and BMH, we have:

a²-2ax+x²=a²+BH²-2a.BH.sen<C and c²-2cy+y²=c²+BH²-2c.BH.sen<A <=>
2ax-x²=2a.BH.sen<C-BH² and 2cy-y²=2c.BH.sen<A-BH².
But 2ax-x²=BA_1.BA_2 and 2cy-y²=BC_2.BC_1.
Then, we have to prove that: 2c.BH.sen<A-BH²=2a.BH.sen<C-BH² <=>2a.sen<C=2c.sen<A (obviously by Sines'law).
Hence, A_1,A_2,C_1,C_2 are concyclics and by the same form, the other 2 quadruples of points are concyclics.
Conclusion: The six points are concyclics!

Have fun!

not_trig Posts: 1799 Location: NORTH CAROLINA!	Posted: Jul 16, 2008, 10:34 am • # 11 What is a sideline!? What is sideline $BC$ ? _________________ that's pretty tricky

kaloi01

Posts: 20
Location: malaysia

Posted: Jul 16, 2008, 10:51 am • # 12

See point $A$ first. let $X$ is the midpoint of $AB$ and $Y$ is the midpoint $AC$ . It follows that $XY \perp AH$ . Thus, $AH$ is the radical axis of circle $B_1B_2H$ and $C_1C_2H$ . It follows that $AC_1*AC_2 = AB_1*AB_2 \Leftrightarrow B_1B_2C_1C_2$ cylcic, by similar argument on $B$ and $C$ we´re done. :lol:

Last edited by kaloi01 on Jul 16, 2008, 11:33 am, edited 1 time in total.

campos

Posts: 403
Location: San Jose, Costa Rica

Posted: Jul 16, 2008, 11:04 am • # 13

it's enough to prove that $O$ is at the same distance of any of the six points... obviously $OA_1 = OA_2$ , so it's enough to prove that the expression $OA_1$ is independent from $A$ . In fact, $OA_1^2 = OM^2 + MA_1^2 = R^2\cos^2A + MH^2$ ...

it can be proven that $MH^2 = R^2(4\cos^2B\cos^2C + \sin^2(B - C))$

so, $OA_1^2 = R^2(\cos^2A + 4\cos^2B\cos^2C + \sin^2(B - C))$ ...

finally, after some work (you can try $\cos A = \sin B\sin C - \cos B\cos C$ ) it follows that it equals $R^2(1-4\cos A\cos B\cos C)$ , and we're done.

Last edited by campos on Jul 17, 2008, 10:21 am, edited 1 time in total.

Virgil Nicula

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Romania

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Posted: Jul 16, 2008, 11:37 am • # 14

orl wrote:

Let $H$ be the orthocenter of an acute triangle $ABC$ . The circle centered at the midpoint of $BC$ and passing through $H$ intersects

the line $BC$ at $A_{1}$ , $A_{2}$ . Similarly define the pairs $B_{1}$ , $B_{2}$ and $C_{1}$ , $C_{2}$ . Prove that $A_{1}$ , $A_{2}$ , $B_{1}$ , $B_{2}$ , $C_{1}$ , $C_{2}$ are concyclically.

Proof. Denote the midpoints $D$ , $E$ , $F$ of the sides $[BC]$ , $[CA]$ , $[AB]$ respectively. Prove easily or is well-known that

$\boxed {\ HA^2 + a^2 = HB^2 + b^2 = HC^2 + c^2 = 4R^2\ }\ (*)$ . Thus, $4\cdot\overline {BA_1}\cdot\overline {BA_2} = 4\cdot\left(BD^2 - HD^2\right) =$ $a^2 - 4\cdot HD^2 =$

$a^2 - 2\cdot\left(HB^2 + HC^2\right) + a^2$ $\stackrel {(*)}{\ \ \implies\ \ }$ $\overline {BA_1}\cdot\overline {BA_2} = \frac 12\cdot \left(a^2 + b^2 + c^2\right) - 4R^2$ $\implies$ $\boxed {\ \overline {BA_1}\cdot\overline {BA_2} = 4R^2\prod\cos A\ }$

because prove easily or is well-known that $\boxed {\ \left(a^2 + b^2 + c^2\right) - 8R^2 = 8 R^2\prod\cos A\ } > 0$ (remark that $\{A_1,A_2\}\subset (BC)$ a.s.o.).

Thus, $\overline {BA_1}\cdot \overline {BA_2} = - \overline {A_1B}\cdot\overline {A_1C} = R^2 - OA_1^2\ \implies\ 4R^2\prod\cos A = ...$ $\implies$ $OA_1^2 = R^2\left(1 - 4\prod\cos A\right)$

(symmetrically in $a$ , $b$ , $c$ ). In conclusion, $\boxed {\ \rho = OA_1 = OA_2 = OB_1 = OB_2 = OC_1 = OC_2 = R\sqrt {1 - 4\cos A\cos B\cos C}\ }$ .

Remark. Since $1 - 8\prod\cos A\ge 0$ obtain easily that $\rho\ge 2R\sqrt {\prod\cos A}$ .

Pohoatza wrote:

I guess it would have been a better IMO problem, at least to avoid the "well-known"-type commentaries.

All right !

Last edited by Virgil Nicula on Jul 16, 2008, 6:47 pm, edited 23 times in total.

Akashnil Posts: 753 Location: Barrackpore, Kolkata	Posted: Jul 16, 2008, 11:44 am • # 15 Let $O$ circumcenter, $D$ midpoint of $BC$ , $N$ nine point center. $OA_1^2=OD^2+DA_1^2=OD^2+DH^2=2(DN^2+ON^2)=const.$

nayel

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Posted: Jul 16, 2008, 12:56 pm • # 16

kaloi01 wrote:

This was exactly my solution.

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gighiuhui Posts: 61	Posted: Jul 16, 2008, 4:02 pm • # 17 I am confused: is IMO a 4.5 hour test or a 4.5 minute test? Someone please help. _________________ Mit '13 pea '09 mop '07 '08 '09 '10

kurt.math Posts: 242	Posted: Jul 16, 2008, 4:41 pm • # 18 gighiuhui wrote: I am confused: is IMO a 4.5 hour test or a 4.5 minute test? Someone please help. I am not sure whether you are being sarcastic It is a 4.5 hour test. But if you are saying the questions are easy...

Albanian Eagle Posts: 1702 Location: Pasadena CA	Posted: Jul 16, 2008, 4:47 pm • # 19 well,... he is sort of right. One could expect these questions in a Putnam exam for example. _________________ Gjergji Zaimi

not_trig

Posts: 1799
Location: NORTH CAROLINA!
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Posted: Jul 16, 2008, 6:13 pm • # 20

As stated, it suffices to show that $OA_1=OA_2=OB_1=\ldots=OC_2$ . Obviously $OA_1=OA_2, OB_1=OB_2$ , etc., as $O$ is on the perpendicular bisector of $A_1A_2$ , etc. So it suffices to show WLOG $OA_1=OB_2$ (assume they both are closer to $C$ .) Then

$OA_1^2 = R^2 - \frac{a^2}{4} + HM_A^2$ , where $M_A$ is the midpoint of $BC$ . (similarly define $M_B,M_C$ .)

So it suffices to show that $HM_A^2 - \frac{a^2}{4} = HM_B^2 - \frac{b^2}{4}$ . But applying the law of cosines to $\triangle M_BCH$ and $M_ACH$ , we have

$HM_A^2 - \frac{a^2}{4} = -(CH)^2 + 2a(CH)\cos \beta$

and

$HM_B^2 - \frac{b^2}{4} = -(CH)^2 + 2b(CH)\cos \alpha$

so they are equal by the law of sines and we're done.

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that's pretty tricky

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