Difference between revisions of "2002 Pan African MO Problems/Problem 5"
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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== | ||
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}} | }} | ||
− | [[Category:Olympiad Geometry Problems]] | + | [[Category: Geometry]][[Category:Olympiad Geometry Problems]] |
Revision as of 20:27, 4 October 2020
Problem
Let 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
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 |