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

Latest revision as of 12:47, 27 May 2024

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.

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

Page

Toolbox

Search

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

Latest revision as of 12:47, 27 May 2024

Contents

Problem

Solution

Solution 2 (by duck_master)

See Also

@@ Line 1: / Line 1: @@
 ==Problem==
-Let <math>\triangle{ABC}</math> 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.
+Let <math>\triangle{ABC}</math> be an acute angled triangle. The circle with diameter <math> AB </math> intersects the sides <math> AC  </math> and <math> BC </math> at points <math> E </math> and <math> F </math> respectively. The tangents drawn to the circle through <math> E </math> and <math> F </math> intersect at <math> P </math>.
-Show that P lies on the altitude through the vertex C.
+Show that <math> P </math> lies on the altitude through the vertex <math> C </math>.
 ==Solution==
@@ Line 34: / Line 34: @@
 draw(e--O--f,dotted);
 </asy>
-Draw lines <math>GA</math> and <math>BH</math>, where <math>G</math> and <math>H</math> are on <math>EP</math> and <math>FP</math>, respectively.  Because <math>GA</math> and <math>GE</math> are [[tangents]] as well as <math>HB</math> and <math>HF</math>, <math>GA = GE</math> and <math>HB = HF</math>.  Additionally, because <math>EP</math> and <math>FP</math> are tangents, <math>EP = FP</math>.
+Draw lines <math>GA</math> and <math>BH</math>, where <math>G</math> and <math>H</math> are on <math>EP</math> and <math>FP</math>, respectively.  Because <math>GA</math> and <math>GE</math> are [[tangent|tangents]] as well as <math>HB</math> and <math>HF</math>, <math>GA = GE</math> and <math>HB = HF</math>.  Additionally, because <math>EP</math> and <math>FP</math> are tangents, <math>EP = FP</math>.
 <br>
@@ Line 62: / Line 62: @@
 <br>
 Thus, by the Base Angle Theorem, <math>\angle PEC = \angle PCE</math>, so <math>\angle PCE = a</math>.  Since <math>\angle GAE = \angle ECP</math>, by the Alternating Interior Angle Converse, <math>GA \parallel CP</math>.  Therefore, since <math>GA \perp AB</math>, <math>CP \perp AB</math>, and <math>P</math> must be on the altitude of <math>\triangle ABC</math> that is through vertex <math>C</math>.
+== 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 <math>D</math> be the [[intersection]] of <math>AF</math> and <math>BE</math>. Note that <math>\angle CED = 180^\circ - \angle AED = 180^\circ - \angle AEB = 90^\circ</math>, and similarly <math>\angle CFD = 180^\circ - \angle BFD = 180^\circ - \angle BFA = 90^\circ</math>. Thusly, <math>CEDF</math> is a [[cyclic quadrilateral]], and <math>CD</math> is the [[diameter]] of its [[circumcircle]].
+Next, let <math>N</math> be the intersection of <math>CD</math> and <math>AB</math>; we claim that <math>CN\perp AB</math>. Note that <math>\angle NCE = \angle DCE = \angle DFE = \angle AFE = \angle ABE = \angle NBE</math>, so <math>NECB</math> is cyclic. Then <math>\angle CNB = \angle CEB = 90^\circ</math>, so <math>CN\perp AB</math>.
+Furthermore, we claim that <math>P</math> is the midpoint of <math>CD</math>. To show this, we use the method of phantom points: we let <math>P'</math> be the midpoint of <math>CD</math>. Then <math>\angle PED = \angle PEB = \angle PEO - \angle OEB = 90^\circ - \angle OBE = 90^\circ - \angle ABE = \angle AEB</math>, and <math>\angle P'ED = \angle P'DE = \angle CDE = \angle CFE = 180^\circ - \angle BFE = \angle AEB</math>. Since the two values match, we have <math>\angle PED = \angle P'ED</math>. Similarly, we show that <math>\angle PFD = \angle P'FD</math>. This necessarily implies <math>P = P'</math>.
+Finally, we show that <math>P</math> lies on the height from <math>C</math> to <math>AB</math>. Since <math>CN\perp AB</math>, we know that <math>CN</math> is the height from <math>C</math> to <math>AB</math>. But <math>CP\parallel CD\parallel CN</math>, so <math>P</math> lies on <math>CN</math> and we are done.
 ==See Also==
@@ Line 69: / Line 135: @@
 |num-a=6
 }}
+[[Category:Olympiad Geometry Problems]]