Problem:
Let be a triangle with orthocenter and let be the second intersection of the circumcircle of triangle with the internal bisector of the angle . Let be the circumcenter of triangle and the orthocenter of triangle . Prove that the length of segment is equal to the circumradius of triangle .
Solution:
It is well-known that the reflection of the orthocenter in the line lies on the circumcircle of triangle . Hence, the circumcenter of triangle coincides with the circumcenter of triangle . But since is the reflection of in the line , the triangles and are symmetric with respect to , and the circumcenter of triangle must be the reflection of the circumcenter of triangle in the line , i. e. the reflection of the circumcenter of triangle in the line .
Now since the quadrilateral is cyclic and since are the orthocenters of triangles , and , respectively, we have that
Hence the point lies on the circumcircle of triangle , and therefore , where denotes the circumradius of triangle .
On the other hand, note that the lines are the perpendicular bisectors of the segments , and , respectively, we get
Thus . Combining this with and with the parallelism of the lines and (note that these two lines are both perpendicular to ), we conclude that the trapezoid is isosceles, and therefore . This completes our proof.
Remark. If is right-angled at , then the statement is trivially true if we convene that the circumcenter of is the midpoint of and that the orthocenter of is the midpoint of . Then, we have that .
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Because is acute, lies inside the triangle. We consider the configuration show above. (For other possible configurations, it is not difficult to adjust our proof properly.)
Let and denote the circumcenters of triangles and respectively. Let and denote the circumcircle and the circumradius of triangle respectively. We will show that
Because and are the circumcenters of triangle and , line is the perpendicular bisector of segment . Because is the orthocenter of triangle . Hence both lines and are perpendicular to line , implying that is a trapezoid with .
Because and are the circumcenters of triangles and , line is the perpendicular bisector of segment . Because and , the acute angles formed by lines and is equal to the acute angle formed by lines and ; that is, . Likewise, we can can show that . Therefore, we have , implying that ; that is, lies on the perpendicular bisector of segment .
Because is the orthocenter of acute triangle . Because is cyclic, we have . Now in obtuse triangle . (This relates to the fact of orthocenter group: if one point is the orthocenter of the triangle formed by the other three points, then any of the four point is the orthocenter of the triangle formed by the other three.) In particular, this means that lies on ; that is, .
Note that in trapezoid , the perpendicular bisectors of the bases and share a common point . Thus, these two bisectors must coincide; that is, is an isosceles trapezoid with , establishing the first part of (??).
To complete our proof, it suffices to show that . Let be the reflection of across line . It is well known that lies (because .) We note that triangle and its circumcenter and triangle and its circumcenter are respective images of each other across line . In particular, we conclude that , completing our proof.
This problem and solutions were suggested by Titu Andreescu and Cosmin Pohoata.
The problems on this page are the property of the MAA's American Mathematics Competitions