Difference between revisions of "2002 Pan African MO Problems/Problem 5"
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==Problem== | ==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== | ==Solution== | ||
Line 34: | Line 34: | ||
draw(e--O--f,dotted); | draw(e--O--f,dotted); | ||
</asy> | </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> | <br> | ||
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<br> | <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>. | 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== | ==See Also== |
Latest revision as of 12:47, 27 May 2024
Problem
Let be an acute angled triangle. The circle with diameter
intersects the sides
and
at points
and
respectively. The tangents drawn to the circle through
and
intersect at
.
Show that
lies on the altitude through the vertex
.
Solution
Draw lines
and
, where
and
are on
and
, respectively. Because
and
are tangents as well as
and
,
and
. Additionally, because
and
are tangents,
.
Let and
. By the Base Angle Theorem,
and
. Additionally, from the property of tangent lines,
,
,
, and
. Thus, by the Angle Addition Postulate,
and
. Thus,
and
, so
. Since the sum of the angles in a quadrilateral is 360 degrees,
. Additionally, by the Vertical Angle Theorem,
and
. Thus,
.
Now we need to prove that
is the center of a circle that passes through
. Extend line
, and draw point
not on
such that
is on the circle with
. By the Triangle Angle Sum Theorem and Base Angle Theorem,
. Additionally, note that
, and since
,
. Thus, by the Base Angle Converse,
. Furthermore,
. Therefore,
is the diameter of the circle, making
the radius of the circle. Since
is a point on the circle,
.
Thus, by the Base Angle Theorem, , so
. Since
, by the Alternating Interior Angle Converse,
. Therefore, since
,
, and
must be on the altitude of
that is through vertex
.
Solution 2 (by duck_master)
Let be the intersection of
and
. Note that
, and similarly
. Thusly,
is a cyclic quadrilateral, and
is the diameter of its circumcircle.
Next, let be the intersection of
and
; we claim that
. Note that
, so
is cyclic. Then
, so
.
Furthermore, we claim that is the midpoint of
. To show this, we use the method of phantom points: we let
be the midpoint of
. Then
, and
. Since the two values match, we have
. Similarly, we show that
. This necessarily implies
.
Finally, we show that lies on the height from
to
. Since
, we know that
is the height from
to
. But
, so
lies on
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 |