Mmm ! A really nice problem ! Please draw a gifure. Let $QH$ meets $CD$ at $M$ and $OQ$ meets $CD$ at $N$ , where $O$ is the center of the great circle. Let line $QO$ , produced , meets the circle $O$ at $E$. Then $OHMN$ is inscribed and : $QM \cdot QH = QN\cdot QO = \frac{1}{2} QN \cdot QE = \frac{1}{2} QC^2 = \frac{1}{2} QH^2$ Hence $QM \cdot QH = \frac{1}{2} QH^2$ or $QM = MH$ Babis

Valentin Vornicu wrote: From a point $Q$ on a circle with diameter $AB$ different from $A$ and $B$, we draw a perpendicular to $AB$, $QH$, where $H$ lies on $AB$. The intersection points of the circle of diameter $AB$ and the circle of center $Q$ and radius $QH$ are $C$ and $D$. Prove that $CD$ bisects $QH$. Actually is rather simple provided inversion is applied... Center $Q$ and radius $QH$. The big circle turns into line $CD$ and viceversa. If $X$ is the other point where $QH$ meets big circle and $Y$ is the point of intersection of $CD$ and $QH$ then it must be $QY \cdot QX = QH^2 \Rightarrow \frac{QY}{QH} = \frac{QH}{QX} = \frac{1}{2}$ Daniel

stergiu wrote: $\frac{1}{2} QN \cdot QE = \frac{1}{2} QC^2$ Don't understand this part, could someone explain me?

M4RI0 wrote: stergiu wrote: $\frac{1}{2} QN \cdot QE = \frac{1}{2} QC^2$ Don't understand this part, could someone explain me? The triangle $ECQ$ is right and $CN$ is its height . In Greece we called it as <<theorem of projections>> and you can prove it by similar triangles.

M4RI0 wrote: stergiu wrote: $\frac{1}{2} QN \cdot QE = \frac{1}{2} QC^2$ Don't understand this part, could someone explain me? The similarity of the triangles gives us this relation. Since $QN$ is the diameter of the first cirlce we have $CN$ perpendicular to $CQ$. Aslo we know that the diagonals of kites are perpendicular to each other which is : the diagonals of a kite $QCOD$ are $QO$ and $CD$ are perpendicular to each other. Therefore we have a similarity of the triangles $QCP$ and $QNC$. In other words, $\frac{QC}{QN}= \frac{QP}{QC}$, quivalent to $QC^2=QN \times QP$ Note that the typos are according to my own figure - $O$ is the centre of the first circle - $Q$ is the centre of the second circle - $N$ is the intersection of point of $QO$ with the first circle - $P$ is the intersection point of $CD$ with $QO$ [Maybe they are the same but i have no time check, sorry] Davron

Let V be the intersection point of OH with circle O and let N be the intersection point of DC and QH. Denote angle QDN with x and angle HDC with y. Since QH=QD we have angleDQH=180-2(x+y) so angleDCH=1/2angleDQH=90-(x+y). ALso we have angleDCV=angleDCH+angleHCV=90-(x+y) + angleHCV so angleHCV=90-(x+y) so this means that HC is angle bisector of angleNCV in triangle NCV and from this we conclude that CN/CV=NH/HV or NH=CN/CV*HV (1) Also we have angleQVC=angleQDC=angleQCD=x and since angleNQC=angleVQC it follows that triangles QNC and QVC are similar so QN/NC=QC/VC or QN=NC/VC*QC (2) Point H is a feet of perpendicular from O to VQ in an triangle VOQ (VO=OQ) so QH=HV. Also QH=QC (radius of circle with radius QH) so we conclude that HV=QC (*) Now, from (1), (2) and (*) we conslude that NH=NQ so it means that CD bisects QH. q.e.d. [/i]

delegat wrote: Let $V$ be the intersection point of QH with circle $O$ - the center of the first circle - and let $N$ be the intersection point of $DC$ and $QH$. Denote $\angle QDN = x$ and $\angle HDC = y$. Since $QH=QD$ we have $\angle DQH=180-2(x+y)$ so $\angle DCH= \frac{1}{2} \angle DQH=90-(x+y)$. Also we have $\angle DCV=angleDCH+ \angle HCV=90-(x+y) + \angle HCV$ , so $\angle HCV=90-(x+y)$ .This means that $HC$ is angle bisector of $\angle NCV$ in triangle $NCV$ and from this we conclude that $\frac{CN}{CV} = \frac{NH}{HV}$ or $NH= \frac{CN}{CV} \cdot HV$ (1) Also we have $\angle QVC= \angle QDC= \angle QCD=x$ and since $\angle NQC= \angle VQC$ , it follows that triangles $QNC$ and $QVC$ are similar so $\frac{QN}{NC}= \frac{QC}{VC}$ or $QN= \frac{NC}{VC} \cdot QC$ (2) Point $H$ is a feet of perpendicular from $O$ to $VQ$ in $\triangle VOQ , (VO=OQ)$ so $QH=HV$ . Also $QH=QC$ (radius of circle with radius $QH$) so we conclude that $HV=QC$ (*) Now, from $(1), (2)$ and (*) we conslude that $NH=NQ$ , which means that $CD$ bisects $QH$. q.e.d. [/i] I simply made a little correction(...typo) to this elementary and nice solution! Babis

stergiu wrote: $QM \cdot QH = \frac{1}{2} QH^2$ or $QM = MH$ Babis It follows from Euler theorem in right triangles Since $QE$ is diameter so triangle QCE is right at $\angle C$ and foot of altitude from $\angle C$ is $N$ So $CQ^2= QN\times QE$

Let us denote the the circle with centre Q and radius QH by W. Let CD intersect QH at M. Extend QH to meet the circle with diameter AB at M'. (Note that since QH is perpendicular to the diameter AB, H is the midpoint of QM'.) Now, consider inversion with respect to the circle W. Circle with diameter AB passes through Q, the centre of W. Therefore that circle inverts into a line. Further, the points C and D are fixed under inversion because they lie on W. This implies that the circle with diameter AB inverts into line CD. Thus the inverse of point M' is M. Now , QM' = 2QH. QM * QM' = (QH) ^ 2 (M is inverse of M') So, QM = 1/2 QH Thus M is midpoint of QH. QED

Please see post 3 , by Daniel Babis

Is it so simple or am I doing something wrong? Let $M$ be the intersection of $CD$ and $QH$. $QD = QC \Rightarrow \angle QDC = \angle QCD = \angle QRC \Rightarrow QH^2 = QC^2 = QM\cdot QR = QM\cdot 2QH \Rightarrow QH = 2QM $

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xeroxia wrote: Is it so simple or am I doing something wrong? Let $M$ be the intersection of $CD$ and $QH$. $QD = QC \Rightarrow \angle QDC = \angle QCD = \angle QRC \Rightarrow QH^2 = QC^2 = QM\cdot QR = QM\cdot 2QH \Rightarrow QH = 2QM $ This solution is correct, very nice and simple too, but probably not easier than the solution with inversion ! This solution has two important points : - To take the whole circle and produce $AH$ - To prove that $QC$ is tangent with the circle $(R,M,C) $ Of course it is not an national or IMO level olympiad problem, it is easy for senior students but is still an ellegant and intreresting problems for juniors. Thank you for sending your solution ! Babis

It is a team selection test problem. I think the proposers missed the easy solution.

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Let $M$ be the midpoint of $QH$, and let $QH$ cut $(AQB), (CHD)$ for a second time at $P,N$ respectively. Now $MQ\cdot MP=(\frac{1}{3} MN)(3 MH)=MN\cdot MH$, so $M$ has equal powers with respect to both circles. Thus $M$ lies on the radical axis $CD$.

xeroxia wrote: It is a team selection test problem. I think the proposers missed the easy solution. Yes ! In my notes I see that it is a problem in some Bulgarian olympiad, probably tst 2004. But it is olympiad problem for some other country too ! Your proof is original and I have not seen it before. Babis