Difference between revisions of "1990 USAMO Problems/Problem 5"
(→Solution 4) |
|||
(15 intermediate revisions by 7 users not shown) | |||
Line 2: | Line 2: | ||
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. | |
− | + | 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> | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We know that <math>BCB'C'</math> is a cyclic quadrilateral. Hence, | ||
+ | |||
+ | <math>HB \cdot HB' = HC \cdot HC' </math> | ||
+ | |||
+ | <math>\implies Pow_{\omega_{1}} = Pow_{\omega_{2}} </math> | ||
+ | |||
+ | <math>\implies HM \cdot HN = HP \cdot HQ </math> | ||
+ | |||
+ | <math>\implies MPNQ </math> is cyclic <math>\raggedright\blacksquare </math>. | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{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}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 00:42, 27 September 2024
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.
Solution 4
We know that is a cyclic quadrilateral. Hence,
is cyclic .
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.