In the following, the point of intersection of two lines $ g$ and $ h$ will be abbreviated as $ g\cap h$. Suppose $ ABC$ is a triangle in which $ \angle A = 90^{\circ}$ and $ \angle B > \angle C$. Let $ O$ be the circumcircle of the triangle $ ABC$. Let $ l_{A}$ and $ l_{B}$ be the tangents to the circle $ O$ at $ A$ and $ B$, respectively. Let $ BC \cap l_{A} = S$ and $ AC \cap l_{B} = D$. Furthermore, let $ AB \cap DS = E$, and let $ CE \cap l_{A} = T$. Denote by $ P$ the foot of the perpendicular from $ E$ on $ l_{A}$. Denote by $ Q$ the point of intersection of the line $ CP$ with the circle $ O$ (different from $ C$). Denote by $ R$ be the point of intersection of the line $ QT$ with the circle $ O$ (different from $ Q$). Finally, define $ U = BR \cap l_{A}$. Prove that \[ \frac {SU \cdot SP}{TU \cdot TP} = \frac {SA^{2}}{TA^{2}}. \]
Problem
Source: Indian Postal Coaching 2005; 18-th Korean Mathematical Olympiad 2005, final round, problem 4
Tags: ratio, geometry, circumcircle, geometry unsolved
28.10.2005 12:19
My Sketch showed that his problem is wrong.
02.11.2005 23:23
Mathx, are you sure that the problem is wrong? A credible friend of min claims that he has proven it. Could you please post the picture (figure) of the counterexample?
08.05.2007 11:46
So, any solution?
30.06.2007 13:48
Or is it really wrong?
30.06.2007 21:08
i suppose the biggest problem here is drawing the figure itself
05.07.2007 06:12
pardesi wrote: i suppose the biggest problem here is drawing the figure itself agree,the pictureis really too complex,It hard too draw it down and make it clear
05.07.2007 17:55
But it can be drawn... how to solve it?
13.07.2007 15:53
But then again... who cares? dg
12.12.2009 13:51
The conclusion is true, here is a way to prove: let the tagent line pass through $ {C}$ neet $ {AB}$ at point $ {E'}$ ,by Pascal Theory we get that :$ {E',D,S}$ are collinear. Thus ,$ {E = SD\cap AB = E'}$, So we get that: $ {EC}$ is tagent to $ {\odot O}$, [1] $ {TA^2 = TR*TQ}$;$ {SA^2 = SB*SC}$ $ {\rightarrow}$ $ {\frac {SA^2}{TA^2} = \frac {SB*SC}{TR*TQ}}$; Let $ {RQ\cap CB = X}$ [2] By Melelaus : $ {\frac {XR}{RT}*\frac {TU}{SU}*\frac {SB}{CX} = 1}$;$ {\frac {XQ}{TQ}*\frac {PT}{PS}*\frac {SC}{XC} = 1}$; [3] EASILY GET: $ {CX*XB = XR*XQ}$; By {[1][2][3]} $ {\Longrightarrow}$ $ {\frac {SP}{TP}*\frac {SU}{TU} = (\frac {XR}{TR}*\frac {SB}{XB})*(\frac {XQ}{QT}*\frac {SC}{XC}) = \frac {SB*SC}{TR*TQ} = \frac {SA^2}{TA^2}}$ QED.
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06.02.2017 12:06
It's very simple! Observe that $SA^2$ = $SB\cdot SC$ and $TA^2$ = $TQ\cdot TR$. Now, simply apply sine law with some angle chasing to get $\frac{SU\cdot SP}{TU\cdot TP}$ = $\frac{SB\cdot SC}{TQ\cdot TR}$. This proves the result.