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independent of the position of D

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Source: 3rd JBMO 1999

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#1 h Jun 16, 2004, 10:27 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with $AB=AC$ . Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$ , and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$ . Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$ ).

Greece

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6555 posts

#2 h Jun 17, 2004, 11:05 AM • 4 Y

Y by Adventure10, Mango247, TigerOnion, and 1 other user

In the above post, Iris Aliaj asked whether somebody understood the following solution (in fact, taken from Kalva):

Kalva's solution wrote:

Solution

We show that M must lie on the line through A parallel to BC.
Let O be the center of the circumcircle of ABD and O' the center of the other circumcircle. Let X be the midpoint of AD. Take coordinates with origin the midpoint of BC, x-axis along BC and y-axis through A. It is sufficient to show that the y-coordinates of O and O' sum to that of A, or equivalently that their mean is the same as the y-coordinate of X. But it is easy to see that the x-coordinate of X less that of O equals the x-coordinate of O' less that of X. Hence the y-coordinate differences must also be equal

Here is a simpler solution:

We will prove that the point M lies on the parallel to BC through A. Once this will be proven, it will obviously follow that the distance from M to BC equals the distance from A to BC, hence the triangles BCM and BCA have an equal altitude, and since the bases are equal, too, the areas are equal; in other words, the area of triangle BCM is equal to the area of triangle BCA, and hence independent of the position of D.

Hence it remains only to show that M lies on the parallel to BC through A.

Since BB' and CC' are diameters of corresponding circles, we have < BDB' = 90° and < CDC' = 90°. Hence, both points B' and C' lie on the perpendicular to BC at D. As a consequence, the points D, B' and C' lie on one line.

Now, in the following I will only consider the case when the point C' lies between the points D and B'. The other cases can be done similarly.

Now, since the points B, A, D and B' lie on one circle, < AB'D = < ABD. On the other hand, since the points C, A, D and C' lie on one circle, < ACD = 180° - < AC'D. Finally, since triangle ABC is isosceles, < ABC = < ACB. Hence, we have

< AB'C' = < AB'D = < ABD = < ABC = < ACB
= < ACD = 180° - < AC'D = < AC'B'.

Thus, the triangle AB'C' is isosceles, too, meaning that AB' = AC'. Therefore, the point A lies on the perpendicular bisector of the segment B'C'. Hence, since M is the midpoint of the segment B'C', the line AM is perpendicular to B'C'. But the line B'C' itself is perpendicular to BC (since both points B' and C' lie on the perpendicular to BC at D); hence, it follows that the line AM is parallel to BC. In other words, the point M lies on the parallel to BC through A. This is what we wanted to show.

This may look like a long proof, but the one you cited and you couldn't understand was actually longer, just the writer had left out many steps.

Darij

This post has been edited 2 times. Last edited by darij grinberg, Oct 30, 2005, 10:21 AM

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Stun

#3 h Dec 26, 2004, 6:26 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I think S(MBC)=S(ABC)

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#4 h Oct 30, 2005, 9:00 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $w_1=C(O_1,r_1),\ w_2=C(O_2,r_2)$ be the circumcircles of the triangles
$ABD,\ ACD$ respectively. Because $AD=2r_1\sin B=2r_2\sin C$ results that
the triangles $ABD,\ ACD$ have the same length $r_1=r_2=\rho$
of the circumradius. I note $x=m(\widehat {ABB'})$ .

$B'D\perp BC,\ C'D\perp BC\Longrightarrow D\in B'C',\ \overline {DB'C'}\perp BC$ .

$\widehat {ABB'}\equiv \widehat {ADB'}\equiv \widehat {ADC'}\equiv \widehat {ACC'}\Longrightarrow AB'=2\rho \sin x,\ AC'=2\rho\sin x\Longrightarrow$

$AB'=AC'\Longrightarrow AM\perp B'C'\Longrightarrow AM\parallel BC\Longrightarrow \sigma [BMC]=\sigma [ABC]$ .

Remark. $\sigma [XYZ]$ is the area of the triangle $XYZ$ .

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#5 h Aug 10, 2009, 5:02 PM • 1 Y

Y by Adventure10

Nice problemand here is my solution

Attachments:

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#6 h Aug 11, 2009, 4:58 AM • 4 Y

Y by Adventure10, Mango247, Subhan222111, mathbetter

It is easy to prove B',C',D are collinear
AB=AC BB'=CC'=AB/sin∠ADB=AC/sin∠AC'C ∠C'CA=∠C'DA=∠ABB' => ⊿ABB'≌⊿ACC' => AB'=AC' => AM⊥C'D
done

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#7 h Aug 31, 2012, 12:05 AM • 2 Y

Y by Adventure10, Mango247

Since $BB'$ is a diameter, $DB'$ is perpendicular to $BD$ . Since $CC'$ is a diameter, $DC'$ is perpendicular to $CD$ . Thus $C'$ , $B'$ , and $D$ are collinear (in that order). We also have $<BAB' = 90$ and $<CAC' = 90$ . Thus $<BAC = <B'AC'$ . Also, $<ABC = 180 - <AB'C = <AB'C'$ . Thus $ABC$ and $AB'C'$ are similar. This means that $AB'C'$ is isosceles, so $AM$ is perpendicular to $B'C'$ , so $AM$ is parallel to $BC$ . Thus the area of $MBC$ equals the area of $ABC$ and hence is constant.

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#8 h Jul 29, 2015, 6:02 AM • 3 Y

Y by efang, huashiliao2020, Adventure10

$[asy] import olympiad; import cse5; pointpen = black; pathpen = black; size(10cm); pair A=dir(90), B=dir(190), C=dir(350), D=B/3+2C/3; D(A--B--C--cycle); D(circumcircle(A,B,D),green); D(circumcircle(A,C,D),green); pair Bp=2*circumcenter(A,B,D)-B, Cp=2*circumcenter(A,C,D)-C, M=(Bp+Cp)/2; D(B--Bp,blue); D(Cp--C,blue); D(Bp--Cp,red); D(Bp--A--Cp,dashed+gray); D(A--foot(A,B,C),dotted+gray); MP("A",D(A),A); MP("B",D(B),B); MP("C",D(C),C); MP("D",D(D),dir(270)); MP("B'",D(Bp),dir(50)); MP("C'",D(Cp),Cp); MP("O_B'",D(B/2+Bp/2),dir(300)); MP("O_C'",D(C/2+Cp/2),dir(30)); MP("M",D(M),dir(180)); MP("X",D(foot(A,B,C)),dir(270)); [/asy]$
$\begin{align*} \intertext{Let $X$ be the foot of the altitude from $A$ to $BC$. Since $BB'$ and $CC'$ are diameters,} \measuredangle BAB'=90^\circ&\text{ and }\measuredangle CAC'=90^\circ\\ \intertext{Thus,} \measuredangle BAB'&=\measuredangle CAC'.\\ \intertext{Adding $\measuredangle B'AC=-\measuredangle CAB'$ to both sides,} \measuredangle BAB'-\measuredangle CAB'&=\measuredangle C'AC-\measuredangle CAB'.\\ \measuredangle BAC&=\measuredangle B'AC'.\\ \intertext{Thus, there is a spiral similarity centered at $A$ with angle $90^\circ$ that maps $B\mapsto B'$, $C\mapsto C'$. Additionally, it maps $X\mapsto M$. Thus, $\measuredangle XAM=90^\circ$, and it does not depend on the choice of $D$. Hence, $AM\parallel BC$, and the height from $M$ to $BC$ is constant. Since $BC$ is fixed, we have}\\[-2em] [MBC]&=\frac{1}{2}BC\cdot \delta(M,BC)\text{ is constant. }\\[-2.4em] \intertext{\hfill$\Box$} \end{align*}$

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KRIS17

#9 h Dec 2, 2018, 10:06 PM • 2 Y

Y by Adventure10, Mango247

Its easy to see that $B'$ , $C'$ , $D$ are collinear (since angle $B'DB$ = $C'DC$ = $90^o$ ).

Applying Sine rule in triangle $ABC$ , we get:
$Sin(BAD) / Sin(CAD) = BD / DC$

Since $BAB'D$ and $CC'AD$ are cyclic quadrilaterals, anlge $(BAD)$ = anlge $(BB'D)$
and $angle (CAD) = angle (CC'D)$

So, $Sin(BB'D) / Sin(CC'D) = BD / DC$

So $BD / Sin(BB'D) = DC / Sin(CC'D)$
Thus, $BB' = CC'$ (the circumcirlcles $\mathcal{C}_1,\mathcal{C}_2$ are congruent).

From right traingles $BB'A$ and $CC'A$ , we have:
$AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}$
So $AC' = AB'$

Since $M$ is the midpoint of $B'C'$ , $AM$ is perpendicular to $B'C'$ and hence $AM$ is parallel to $BC$ .

So area of traiangle $MBC$ = area of traingle $ABC$ and hence is independent of position of $D$ on $BC$ .

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#10 h Oct 20, 2024, 12:54 PM

Y by

I think we can solve faster using your method; if we use a well-known property of the radical axis AD it is obvious that AD is the perpendicular bisector of O_1O_2, where O_1 and O_2 are centers of c1 and c2, respectively. Hence, triangle AO_1O_2 is isosceles with AO_1 = AO_2

Therefore, c1 and c2 are congruent as their radii are equal. So BB' = CC'.

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