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Difference between revisions of "2002 Pan African MO Problems/Problem 5"

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[[Category:Olympiad Geometry Problems]]

Revision as of 22:53, 6 January 2020

Problem

Let $\triangle{ABC}$ be an acute angled triangle. The circle with diameter AB intersects the sides AC and BC at points E and F respectively. The tangents drawn to the circle through E and F intersect at P. Show that P lies on the altitude through the vertex C.

Solution

[asy] pair a=(-65,0),O=(0,0),b=(65,0),e=(-39,52),f=(25,60),c=(-9.286,111.429),p=(-9.286,74.286),g=(-65,32.5),h=(65,43.333); draw(arc(O,b,a,CCW)); draw(a--b--c--a); dot(a); label("$A$",a,SW); dot(b); label("$B$",b,SE); dot(c); label("$C$",c,N); dot(e); label("$E$",e,NW); dot(f); label("$F$",f,NE); dot(O); label("$O$",O,S); dot(p); label("$P$",p,NE); dot(g); label("$G$",g,NW); dot(h); label("$H$",h,NE); draw(a--g--p--(65,43.333)--b,dotted); draw(c--p,dotted); draw(e--O--f,dotted); [/asy] Draw lines $GA$ and $BH$, where $G$ and $H$ are on $EP$ and $FP$, respectively. Because $GA$ and $GE$ are tangents as well as $HB$ and $HF$, $GA = GE$ and $HB = HF$. Additionally, because $EP$ and $FP$ are tangents, $EP = FP$.


Let $\angle GAE = a$ and $\angle HBF = b$. By the Base Angle Theorem, $\angle GEA = a$ and $\angle HFB = b$. Additionally, from the property of tangent lines, $GA \perp AO$, $GP \perp EO$, $PH \perp FO$, and $HB \perp BO$. Thus, by the Angle Addition Postulate, $\angle OEA = \angle OAE = 90-a$ and $\angle OBF = \angle OFB = 90-b$. Thus, $\angle EOA = 2a$ and $\angle FOB = 2b$, so $\angle EOF = 180-2a-2b$. Since the sum of the angles in a quadrilateral is 360 degrees, $\angle EPF = 2a+2b$. Additionally, by the Vertical Angle Theorem, $\angle GEA = \angle PEC = a$ and $\angle HFB = \angle PFC = b$. Thus, $\angle ECF = a+b$. [asy] pair e=(-39,52),f=(25,60),c=(-9.286,111.429),p=(-9.286,74.286),g=(-65,32.5),h=(65,43.333),j=(-43.572,88.572); draw(c--p,dotted); draw(e--c--f); draw(circle(p,37.143)); draw(p--e,dotted); draw(p--f,dotted); draw(p--j--e,dotted); dot(e); label("$E$",e,SW); dot(f); label("$F$",f,SE); dot(p); label("$P$",p,S); dot(c); label("$C$",c,N); dot(j); label("$J$",j,NW); [/asy] Now we need to prove that $P$ is the center of a circle that passes through $C, E, F$. Extend line $PF$, and draw point $J$ not on $F$ such that $J$ is on the circle with $C, E, F$. By the Triangle Angle Sum Theorem and Base Angle Theorem, $\angle PEF = \angle PFE = \tfrac12 \cdot (180 - \angle EPF) = 90 - \angle ECF$. Additionally, note that $\angle EPJ = 180-\angle EPF = 180 - 2 \angle ECF$, and since $\angle EJF = \angle ECF$, $\angle JEP = \angle EJP$. Thus, by the Base Angle Converse, $PJ = PE$. Furthermore, $\angle JEP + \angle PEF = 90 - \angle ECF + \angle ECF = 90^\circ$. Therefore, $JF$ is the diameter of the circle, making $PF$ the radius of the circle. Since $C$ is a point on the circle, $PF = PC$.


Thus, by the Base Angle Theorem, $\angle PEC = \angle PCE$, so $\angle PCE = a$. Since $\angle GAE = \angle ECP$, by the Alternating Interior Angle Converse, $GA \parallel CP$. Therefore, since $GA \perp AB$, $CP \perp AB$, and $P$ must be on the altitude of $\triangle ABC$ that is through vertex $C$.

See Also

2002 Pan African MO (Problems)
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All Pan African MO Problems and Solutions
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