Prove that SD = SM - Poland 2003 Third Round

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Prove that SD = SM - Poland 2003 Third Round

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#1 h Nov 1, 2010, 7:30 PM • 2 Y

Y by Adventure10, Mango247

In an acute-angled triangle $ABC, CD$ is the altitude. A line through the midpoint $M$ of side $AB$ meets the rays $CA$ and $CB$ at $K$ and $L$ respectively such that $CK = CL.$ Point $S$ is the circumcenter of the triangle $CKL.$ Prove that $SD = SM.$

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#2 h Nov 1, 2010, 8:19 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Recently posted in the current geometry marathon with similar notations.

barcelona wrote:

It is given $\triangle ABC$ and $M$ is the midpoint of $AB.$ Let $\ell$ pass through $M.$ $K \equiv \ell \cap AC$ and $L \equiv \ell \cap BC,$ such that $CK=CL.$ Let $CD \perp AB,$ $D\in AB$ and $O$ is the center of the circumcircle of $\triangle CKL.$ Prove that $OM=OD.$

Since $MA = MB$ and $CK = CL,$ by Menelaus' theorem for $\triangle ABC$ cut by the straight line $KML,$ we get $AK = BL.$ Thus, $E \equiv \odot(CKL) \cap \odot(ABC) \not\equiv C$ is the center of the rotation carrying $AK$ into $BL,$ and due to $EA = EB,$ then $E$ is the midpoint of the arc $AB$ of $\odot(ABC).$ Obviously, $E$ lies on the internal bisector of $\angle ACB$ $\Longrightarrow$ $O \in CE.$ Since $CE$ is the circumdiameter of $\triangle CKL$ perpendicular to $KL,$ it follows that $O$ is the midpoint of $CE.$ Moreover, if $A',B'$ denote the midpoints of $CB,CA,$ then $O$ is the intersection of $CE$ with the perpendicular bisector of segment $A'B',$ since the positive homothety with center $C$ and factor $\frac {_1}{^2}$ transforms $A,B$ into $A',B',$ $E$ into $O$ and the perpendicular bisector $EM$ of $AB$ into the perpendicular bisector of $A'B'.$ Now, it remains to see that quadrilateral $A'B'DM$ is an isosceles trapezoid, in which $O$ lies on its symmetry axis. Indeed, $\triangle MA'B'$ and $\triangle DB'A'$ are symmetric about the perpendicular bisector of $A'B',$ hence $OD = OM.$

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jgnr

#3 h Nov 3, 2010, 12:05 PM • 2 Y

Y by Adventure10, Mango247

Let the line through $B$ parallel to $CA$ meet line $KL$ at $P$ . So $\angle BPL=\angle LKC=\angle CLK=\angle BLP$ , so $BL=BP$ . We also have $\triangle AMK\cong\triangle BMP$ , so $BP=AK$ , therefore $AK=BL$ . WLOG assume $a\ge b$ , so $AK=BL=\frac{a-b}2$ . Let $X,Y,Z$ be the projections of $S$ onto $AB,BC,CA$ respectively. We get $CY=\frac{a+b}4$ , $BY=\frac{3a-b}4$ , $CZ=\frac{a+b}4$ , $ZA=\frac{3b-a}4$ . Note that $CY^2+BX^2+AZ^2=BY^2+AX^2+CZ^2$ , so $BX^2-AX^2=BY^2-AZ^2$ $\implies$ $(BX-AX)(BX+AX)=\frac{9a^2-6ab+b^2}{16}-\frac{9b^2-6ab+a^2}{16}$ $\implies$ $(2BX-c)c=\frac{a^2-b^2}2$ . Therefore $BX=\frac{2c^2+a^2-b^2}{4c}$ . Also note that $BM+BD=\frac{c}2+\frac{a^2+c^2-b^2}{2c}=\frac{2c^2+a^2-b^2}{2c}=\frac{BX}2$ . So $X$ is the midpoint of $MD$ and hence $SD=SM$ , as desired.

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jayme

#4 h Nov 3, 2010, 2:01 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
1. (0) the circumcircle of ABC and O its center
2. (1) the circumcircle of CKL and O1 its center
3. (2) the circumcircle of ADM and O2 its center
4. E the second point of intersection of (1) and (2)
5. P, N the second points of intersection of AO1 with (1), (2) resp.
6. K, L the points of intersection of MN with CA, CB resp.

7. P, M and E are collinear (by a converse of the Reim's theorem)
8. O1O2 // PM (by the midpoint theorem)
9. If we prove that (0) goes through P, then O is on PME and O1O2 perpendicular to DM
10. O1O2 being the mediator of DM, we are done.
Sincerely
Jean-Louis

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