Difference between revisions of "1990 USAMO Problems/Problem 5"
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An acute-angled triangle <math>ABC</math> is given in the plane. The circle with diameter <math>\, AB \,</math> intersects altitude <math>\, CC' \,</math> and its extension at points <math>\, M \,</math> and <math>\, N \,</math>, and the circle with diameter <math>\, AC \,</math> intersects altitude <math>\, BB' \,</math> and its extensions at <math>\, P \,</math> and <math>\, Q \,</math>. Prove that the points <math>\, M, N, P, Q \,</math> lie on a common circle. | An acute-angled triangle <math>ABC</math> is given in the plane. The circle with diameter <math>\, AB \,</math> intersects altitude <math>\, CC' \,</math> and its extension at points <math>\, M \,</math> and <math>\, N \,</math>, and the circle with diameter <math>\, AC \,</math> intersects altitude <math>\, BB' \,</math> and its extensions at <math>\, P \,</math> and <math>\, Q \,</math>. Prove that the points <math>\, M, N, P, Q \,</math> lie on a common circle. | ||
− | == Solution == | + | == Solution 1== |
Let <math>A'</math> be the intersection of the two circles (other than <math>A</math>). <math>AA'</math> is perpendicular to both <math>BA'</math>, <math>CA'</math> implying <math>B</math>, <math>C</math>, <math>A'</math> are collinear. Since <math>A'</math> is the foot of the altitude from <math>A</math>: <math>A</math>, <math>H</math>, <math>A'</math> are concurrent, where <math>H</math> is the orthocentre. | Let <math>A'</math> be the intersection of the two circles (other than <math>A</math>). <math>AA'</math> is perpendicular to both <math>BA'</math>, <math>CA'</math> implying <math>B</math>, <math>C</math>, <math>A'</math> are collinear. Since <math>A'</math> is the foot of the altitude from <math>A</math>: <math>A</math>, <math>H</math>, <math>A'</math> are concurrent, where <math>H</math> is the orthocentre. | ||
Now, <math>H</math> is also the intersection of <math>BB'</math>, <math>CC'</math> which means that <math>AA'</math>, <math>MN</math>, <math>PQ</math> are concurrent. Since <math>A</math>, <math>M</math>, <math>N</math>, <math>A'</math> and <math>A</math>, <math>P</math>, <math>Q</math>, <math>A'</math> are cyclic, <math>M</math>, <math>N</math>, <math>P</math>, <math>Q</math> are cyclic by the radical axis theorem. | Now, <math>H</math> is also the intersection of <math>BB'</math>, <math>CC'</math> which means that <math>AA'</math>, <math>MN</math>, <math>PQ</math> are concurrent. Since <math>A</math>, <math>M</math>, <math>N</math>, <math>A'</math> and <math>A</math>, <math>P</math>, <math>Q</math>, <math>A'</math> are cyclic, <math>M</math>, <math>N</math>, <math>P</math>, <math>Q</math> are cyclic by the radical axis theorem. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Define <math>A'</math> as the foot of the altitude from <math>A</math> to <math>BC</math>. Then, <math>AA' \cap BB' \cap CC'</math> is the orthocenter. We will denote this point as <math>H</math>. | ||
+ | Since <math>\angle AA'C</math> and <math>\angle AA'B</math> are both <math>90^{\circ}</math>, <math>A'</math> lies on the circles with diameters <math>AC</math> and <math>AB</math>. | ||
+ | |||
+ | Now we use the Power of a Point theorem with respect to point <math>H</math>. From the circle with diameter <math>AB</math> we get <math>AH \cdot A'H = MH \cdot NH</math>. From the circle with diameter <math>AC</math> we get <math>AH \cdot A'H = PH \cdot QH</math>. Thus, we conclude that <math>PH \cdot QH = MH \cdot NH</math>, which implies that <math>P</math>, <math>Q</math>, <math>M</math>, and <math>N</math> all lie on a circle. | ||
+ | |||
+ | ==Solution 3 (Radical Lemma)== | ||
+ | |||
+ | Let <math>\omega_1</math> be the circumcircle with diameter <math>AB</math> and <math>\omega_2</math> be the circumcircle with diameter <math>AC</math>. We claim that the second intersection of <math>\omega_1</math> and <math>\omega_2</math> other than <math>A</math> is <math>A'</math>, where <math>A'</math> is the feet of the perpendicular from <math>A</math> to segment <math>BC</math>. Note that <cmath>\angle AA'B=90^{\circ}=\angle AB'B</cmath> so <math>A'</math> lies on <math>\omega_1.</math> Similarly, <math>A'</math> lies on <math>\omega_2</math>. Hence, <math>AA'</math> is the radical axis of <math>\omega_1</math> and <math>\omega_2</math>. By the Radical Lemma, it suffices to prove that the intersection of lines <math>MN</math> and <math>PQ</math> lie on <math>AA'</math>. But, <math>MN</math> is the same line as <math>CC'</math> and <math>PQ</math> is the same line as <math>BB'</math>. Since <math>AA', BB'</math>, and <math>CC'</math> intersect at the orthocenter <math>H</math>, <math>H</math> lies on the radical axis <math>AA'</math> and we are done. <math>\blacksquare</math> | ||
{{alternate solutions}} | {{alternate solutions}} | ||
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== See Also == | == See Also == | ||
− | {{USAMO box|year=1990|num-b=4|after= | + | {{USAMO box|year=1990|num-b=4|after=Last Question}} |
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356630#356630 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356630#356630 Discussion on AoPS/MathLinks] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 20:52, 24 October 2018
Problem
An acute-angled triangle is given in the plane. The circle with diameter
intersects altitude
and its extension at points
and
, and the circle with diameter
intersects altitude
and its extensions at
and
. Prove that the points
lie on a common circle.
Solution 1
Let be the intersection of the two circles (other than
).
is perpendicular to both
,
implying
,
,
are collinear. Since
is the foot of the altitude from
:
,
,
are concurrent, where
is the orthocentre.
Now, is also the intersection of
,
which means that
,
,
are concurrent. Since
,
,
,
and
,
,
,
are cyclic,
,
,
,
are cyclic by the radical axis theorem.
Solution 2
Define as the foot of the altitude from
to
. Then,
is the orthocenter. We will denote this point as
.
Since
and
are both
,
lies on the circles with diameters
and
.
Now we use the Power of a Point theorem with respect to point . From the circle with diameter
we get
. From the circle with diameter
we get
. Thus, we conclude that
, which implies that
,
,
, and
all lie on a circle.
Solution 3 (Radical Lemma)
Let be the circumcircle with diameter
and
be the circumcircle with diameter
. We claim that the second intersection of
and
other than
is
, where
is the feet of the perpendicular from
to segment
. Note that
so
lies on
Similarly,
lies on
. Hence,
is the radical axis of
and
. By the Radical Lemma, it suffices to prove that the intersection of lines
and
lie on
. But,
is the same line as
and
is the same line as
. Since
, and
intersect at the orthocenter
,
lies on the radical axis
and we are done.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1990 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.