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staretak Posts: 62	Posted: Mar 23, 2008, 7:21 pm • # 1 Let ABC be a triangle and let AA' be a diameter of its circumcircle (O). The tangents at B,C of (O) intersect at P. If PS, PT are perpendicular to AB, AC respectively, show that $\angle{A'SP} = \angle{A'TP}$

Nemion

Posts: 177
Location: Caradium

Posted: Mar 23, 2008, 8:50 pm • # 2

$BA{'}{||}SP$ and $CA{'}{||}PT$ then $\angle{A{'}SP}=\angle{SA{'}B}$ analogously $\angle{A{'}TP}=\angle{TA{'}B}$

$BS=\cos{C}(BP)$ and $BA{'}=\cos{C}(2R)$ then, $\frac{BS}{BA{'}}=\frac{BP}{2R}$ analogously $\frac{CT}{CA{'}}=\frac{BP}{2R}$ then $BA{'}S$ is similar to $CA{'}T$ because $\angle{SBA{'}}=\angle{TCA{'}}=\frac{\pi}{2}$ (and the ratios) then $\angle{BA{'}S}=\angle{CA{'}T}$ and therefore $<A{'}SP=<A{'}TP$

staretak

Posts: 62

Posted: Mar 24, 2008, 7:27 am • # 3

Nice solution...

I will present another synthetic solution (without the use of ratios etc.)

Consider points Q,R such that $BQ//=A'R \perp AA'$ and $BQ=A'R=BP(=CP)$ then $BQRA'$ is parallelogram so $QR \perp BA$ and if $QR \cap BA=G$ then $AGA'R$ is cyclic so $\angle RAA'= \angle RGA'= \angle GA'B =\angle BA'S$ (since $BQ$ is the reflection of $BP$ across $BC$ ). Similarly, $\angle CA'T=\angle RAA'$ so $\angle BA'S=\angle CA'T$ .

I posted this exercise here, not because it is difficult or something but because it is useful for a simple solution(without polar theory) to another exercise (which was posted by Silouan and was given to the Bulgarian team):

Let $M,N$ be two interior points in a circle $C(O,R)$ such that $O$ is the mid point of $MN$ . If $S$ is a point that lies on the circle and $E,F$ the intersection points of the lines $SM, SN$ with the circle then prove that the if the tangents at $E,F$ meet at $I$ then the perpendicular bisector of $MN$ goes through the mid point of $SI$ .

The QuattoMaster 6000

Posts: 1196
Location: California
India

Posted: Mar 25, 2008, 7:57 pm • # 4

staretak wrote:

Let ABC be a triangle and let AA' be a diameter of its circumcircle (O). The tangents at B,C of (O) intersect at P. If PS, PT are perpendicular to AB, AC respectively, show that $\angle{A'SP} = \angle{A'TP}$

Nice problem!
Alternate Solution

Let $N$ be the midpoint of $A'B$ and $Q$ be that of $A'C$ . Thus, $\angle NOB = \frac {\angle BOA'}{2} = \angle BAA'$ . Also, $\angle A'BA = 90 = \angle ONB$ since $N$ is the midpoint of $A'B$ , so $\triangle ONB\sim \triangle ABA'$ . Also, $\angle PBS = 180 - \angle ABC - \angle PBC = 180 - \angle ABC - \angle BAC = \angle ACB = \angle AA'B$ . Since $\angle PSB = 90 = \angle ABA'$ , we see that $\triangle PSB\sim \triangle ABA'\sim \triangle ONB\implies \triangle BSN\sim \triangle BPO$ . Similarly, $\triangle CQT\sim \triangle COP\cong \triangle BPO\sim \triangle BSN$ . Since $N$ and $Q$ are the midpoints of $A'B$ and $A'C$ , we see that $\triangle BA'S\sim \triangle CA'T\implies \angle CTA' = \angle BSA'\implies 90 - \angle BSA' = 90 - \angle CTA'\implies \angl...$

Virgil Nicula

Posts: 4868
Romania

Blog: View Blog

Posted: Mar 29, 2008, 8:42 am • # 5

staretak wrote:

Let $ABC$ be a triangle and let $AA'$ be a diameter of its circumcircle $w = C(O)$ . The tangents at $B$ , $C$ of $w$ intersect at $P$ . If $PS$ , $PT$ are perpendicular to $AB$ , $AC$ respectively show that $\widehat {A'SP}\equiv\widehat {A'TP}$ .

Proof. Prove easily that the following problem ("any similarity of the notations is absolutely random") :

Quote:

Let $A'$ be the orthocenter of the triangle $AUV$ . Denote $\left\{\begin{array}{c} B\in AU\cap VA' \\\ C\in AV\cap UA'\end{array}$ and the middlepoints $P$ , $S$ , $T$ of the segments $[UV]$ , $[UB]$ , $[VC]$ respectively. Then $\left\{\begin{array}{c} PS\perp AB \\\ PT\perp AC\end{array}$ , the lines $PB$ , $PC$ are the tangents from the point $P$ to the circumcircle of the triangle $ABC$ and $\widehat {A'SP}\equiv\widehat {A'TP}$ .

If we"ll erase the points $U$ , $V$ , then we"ll obtain the proposed problem !

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Geometry of Triangle

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