Difference between revisions of "2005 USAMO Problems/Problem 3"
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== Solution == | == Solution == | ||
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+ | === Soltuion 1 === | ||
Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>. It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>. All our angles will be directed, and measured mod <math>\pi</math>. | Let <math>B_1'</math> be the second intersection of the line <math>C_1A</math> with the circumcircle of <math>APC</math>, and let <math>Q'</math> be the second intersection of the circumcircle of <math>B_1' C_1P</math> and line <math>BC</math>. It is enough to show that <math>B_1'=B_1</math> and <math>Q' =Q</math>. All our angles will be directed, and measured mod <math>\pi</math>. | ||
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It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>. Hence <math>Q' = Q</math>. Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>. Hence <math>B_1' = B_1</math>, as desired. <math>\blacksquare</math> | It follows that <math>Q'</math> is the intersection of <math>BC</math> and the line parallel to <math>AC</math> and passing through <math>C_1</math>. Hence <math>Q' = Q</math>. Then <math>B_1'</math> is the second intersection of the circumcircle of <math>APC</math> and the line parallel to <math>AB</math> passing through <math>Q</math>. Hence <math>B_1' = B_1</math>, as desired. <math>\blacksquare</math> | ||
− | ==Solution 2== | + | ===Solution 2=== |
''Lemma''. <math>B_1, A, C_1</math> are collinear. | ''Lemma''. <math>B_1, A, C_1</math> are collinear. | ||
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As a result, we have <math>\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ</math>, so <math>B_1 C_1 PQ</math> is cyclic, as desired. | As a result, we have <math>\angle C_1 B_1 Q = \angle C_1 AB = \angle C_1 PB = 180^\circ - \angle C_1 PQ</math>, so <math>B_1 C_1 PQ</math> is cyclic, as desired. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Due to the parallel lines, <math>m\angle C_1QB_1=m\angle A.</math> Therefore, it suffices to prove that <cmath>m\angle C_1PB_1=m\angle C_1BA+m\angle ACB_1=m\angle A.</cmath> Note that <math>m\angle BC_1A+m\angle AB_1C=180^{\circ}</math> by the cyclic quadrilaterals. Now, the condition simplifies to proving <math>C_1,A,B_1</math> collinear. | ||
+ | |||
+ | Use barycentric coordinates. Let <cmath>A=(1,0,0), B=(0,1,0), C=(0,0,1), P=(0,p,1-p), Q=(0,q,1-q).</cmath> By the parallel lines, <cmath>C_1=(1-q-z',q,z'), B_1=(q-y',y',1-q)</cmath> for some <math>y',z'.</math> As <math>A</math> is a vertex of the reference triangle, we must prove that <math>q(1-q)=y'z'.</math> | ||
+ | |||
+ | Now we find the circumcircle of <math>\triangle APC.</math> Let its equation be <cmath>-a^2yz-b^2xz-c^2xy+(x+y+z)(ux+vy+wz)=0.</cmath> Substituting in <math>A,C</math> gives <math>u=w=0.</math> Substituting in <math>P</math> yields <math>-a^2p(1-p)+vp=0\implies v = a^2(1-p).</math> Substituting in <math>B_1</math> yields | ||
+ | <cmath>0=-a^2y'(1-q)-b^2(q-y')(1-q)-c^2y'(q-y')+a^2(1-p)y'=0.</cmath> | ||
+ | <cmath>c^2y'^2+(a^2q-a^2p+b^2(1-q)-c^2q)y'-b^2q(1-q)=0</cmath> | ||
+ | |||
+ | Similarly, the circumcircle of <math>\triangle APB</math> is <cmath>-a^2yz-b^2xz-c^2xy+(x+y+z)a^2pz</cmath> | ||
+ | |||
+ | <cmath>0=-a^2qz'-b^2z'(1-q-z')-c^2q(1-q-z')+a^2pz'=0</cmath> | ||
+ | <cmath>b^2 z'^2+(a^2p-a^2q-b^2(1-q)+c^2q)z'-c^2q(1-q)=0</cmath> | ||
+ | |||
+ | Let <math>k=a^2p-a^2q-b^2(1-q)+c^2q,</math> and note that both <math>y',z'</math> are negative. By the quadratic formula, we have | ||
+ | <cmath>y'=\frac{k-\sqrt{k^2+4b^2c^2q(1-q)}}{2c^2}, z'=\frac{-k-\sqrt{k^2+4b^2c^2q(1-q)}}{2b^2}</cmath> | ||
+ | Multiplying these two, we have <cmath>y'z'=\frac{4b^2c^2q(1-q)}{4b^2c^2}=q(1-q),</cmath> as desired. <math>\blacksquare</math> | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 08:47, 29 May 2016
Problem
(Zuming Feng) Let be an acute-angled triangle, and let
and
be two points on side
. Construct point
in such a way that convex quadrilateral
is cyclic,
, and
and
lie on opposite sides of line
. Construct point
in such a way that convex quadrilateral
is cyclic,
, and
and
lie on opposite sides of line
. Prove that points
, and
lie on a circle.
Solution
Soltuion 1
Let be the second intersection of the line
with the circumcircle of
, and let
be the second intersection of the circumcircle of
and line
. It is enough to show that
and
. All our angles will be directed, and measured mod
.
Since points are concyclic and points
are collinear, it follows that
But since points
are concyclic,
It follows that lines
and
are parallel. If we exchange
with
and
with
in this argument, we see that lines
and
are likewise parallel.
It follows that is the intersection of
and the line parallel to
and passing through
. Hence
. Then
is the second intersection of the circumcircle of
and the line parallel to
passing through
. Hence
, as desired.
Solution 2
Lemma. are collinear.
Suppose they are not collinear. Let line intersect circle
(i.e. the circumcircle of
) again at
distinct from
. Because
, we have that
is cyclic. Hence
, so
. Then
must be the other intersection of the parallel to
through
with circle
. Then
is on segment
, so
is contained in triangle
. However, line
intersects this triangle only at point
because
lies on arc
not containing
of circle
, a contradiction. Hence,
are collinear, as desired.
As a result, we have , so
is cyclic, as desired.
Solution 3
Due to the parallel lines, Therefore, it suffices to prove that
Note that
by the cyclic quadrilaterals. Now, the condition simplifies to proving
collinear.
Use barycentric coordinates. Let By the parallel lines,
for some
As
is a vertex of the reference triangle, we must prove that
Now we find the circumcircle of Let its equation be
Substituting in
gives
Substituting in
yields
Substituting in
yields
Similarly, the circumcircle of is
Let and note that both
are negative. By the quadratic formula, we have
Multiplying these two, we have
as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
- <url>Forum/viewtopic.php?p=213011#213011 Discussion on AoPS/MathLinks</url>
2005 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.