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2002 Pan African MO Problems/Problem 5

Revision as of 21:27, 4 October 2020 by Duck master (talk | contribs) (added novel solution to page)

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 $C$.

Solution 2 (by duck_master)

[asy] import graph; pair A, B, O; path circleAB; A = (-5, 0); B = (5, 0); O = (A + B)/2; circleAB = Circle(O, 5); dot(A); dot(B); dot(O); label("$A$", A, SW); label("$O$", O, S); label("$B$", B, SE); draw(A--B); draw(circleAB); pair C, E, F, D; C = (1, 8); E = intersectionpoint(C--A, circleAB); F = intersectionpoint(C--(0.9*B + 0.1*C), circleAB); D = intersectionpoint(A--F, B--E); dot(C); dot(E); dot(F); dot(D); label("$C$", C, NE); label("$E$", E, NW); label("$F$", F, NE); label("$D$", D, SE, blue); draw(A--C--B); draw(A--F--E--B, blue); draw(E--O--F, blue); draw(Circle((C + D)/2, length(C - D)/2), darkgreen); pair Nextend, Npt; Nextend = 2.5*D - 1.5*C; Npt = intersectionpoint(C--Nextend, A--B); dot(Npt, blue); label("$N$", Npt, SE, blue); draw(C--Nextend, blue); pair Etangplus, Etangminus, Ftangplus, Ftangminus, P; Etangplus = E + 2*(E - O)*dir(90); Etangminus = E - 2*(E - O)*dir(90); Ftangplus = F + 2*(F - O)*dir(90); Ftangminus = F - 2*(F - O)*dir(90); P = intersectionpoint(Etangminus -- Etangplus, Ftangminus -- Ftangplus); dot(P); label("$P = P'$", P, NE); draw(Etangminus -- Etangplus); draw(Ftangminus -- Ftangplus); [/asy]

Let $D$ be the intersection of $AF$ and $BE$. Note that $\angle CED = 180^\circ - \angle AED = 180^\circ - \angle AEB = 90^\circ$, and similarly $\angle CFD = 180^\circ - \angle BFD = 180^\circ - \angle BFA = 90^\circ$. Thusly, $CEDF$ is a cyclic quadrilateral, and $CD$ is the diameter of its circumcircle.

Next, let $N$ be the intersection of $CD$ and $AB$; we claim that $CN\perp AB$. Note that $\angle NCE = \angle DCE = \angle DFE = \angle AFE = \angle ABE = \angle NBE$, so $NECB$ is cyclic. Then $\angle CNB = \angle CEB = 90^\circ$, so $CN\perp AB$.

Furthermore, we claim that $P$ is the midpoint of $CD$. To show this, we use the method of phantom points: we let $P'$ be the midpoint of $CD$. Then $\angle PED = \angle PEB = \angle PEO - \angle OEB = 90^\circ - \angle OBE = 90^\circ - \angle ABE = \angle AEB$, and $\angle P'ED = \angle P'DE = \angle CDE = \angle CFE = 180^\circ - \angle BFE = \angle AEB$. Since the two values match, we have $\angle PED = \angle P'ED$. Similarly, we show that $\angle PFD = \angle P'FD$. This necessarily implies $P = P'$.

Finally, we show that $P$ lies on the height from $C$ to $AB$. Since $CN\perp AB$, we know that $CN$ is the height from $C$ to $AB$. But $CP\parallel CD\parallel CN$, so $P$ lies on $CN$ and we are done.

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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