Heres my solution: We already know that angle A > angle B > angle C. Let <OBC = <OCB = x, <OAB = <OBA = y, and <OAC = <OCA = z. Then since BC > AC > AB, z + y > x + y > x + z. However, for OI to intersect AC, <OAC must be greater than <OAB (since AI bisects <CAB). This means that z > y, a contradiction. Thus OI intersects only AB and BC.

froufrou4189 wrote: Heres my solution: However, for OI to intersect AC, <OAC must be greater than <OAB (since AI bisects <CAB). This means that z > y, a contradiction. I don't quite understand this part. Why must <OAC be greater than <OAB? I believe it is possible to construct such a "contradiction"

For OI to intersect AC, O must be on the same side of AI as B. Since AI bisects angle A, angle OAB must be smaller than angle OAC (this means that O is on the same side of AI as B). A similar way to prove this would be to use the above reasoning to conclude that O is on the same side of AI as C and on the same side of CI as B. So if AI intersects BC at D, then O must lie inside triangle CDI, which means it cannot intersect AC.

Here's my solution, very different from the above two: First we show that if given a triangle $ ABC$ with circumcenter $ O$ and incenter $ I$, it is true that $ OI$ intersects a vertex, then the triangle is isosceles. WLOG let that vertex is $ A$. Then $ AI$ is the bisector of $ \angle BAC$ and $ AI$ is the same as $ AO$, so $ \angle BAO\cong\angle CAO$. Since $ \triangle BAO$ and $ \triangle CAO$ are isosceles with $ AO = BO = CO$, it follows that they are congruent and that $ AB = AC$ and $ \triangle ABC$ is isosceles as desired. Say we have $ \triangle ABC$ with circumcenter $ O$ and incenter $ I$. If we move $ A$, $ B$, and $ C$ continuously without allowing the triangle to become isosceles, then $ O$ and $ I$ move continuously because they are solutions to quadratic equations in terms of $ A$, $ B$, and $ C$, and thus the intersection of $ OI$ with the boundary of $ \triangle ABC$ will move continuously. If one of the intersections moves from one side to another, it must pass a vertex, but this would make the triangle become isosceles, so the intersections of $ OI$ with the triangle must remain on the same segments. Also note that if the triangle is scalene then $ O$ and $ I$ must both lie inside the triangle and not coincide, and their line's intersection with the interior and boundary of the triangle should be a line segment with two endpoints because the triangle is convex. We note that in a triangle with sides $ AB = 5$, $ AC = 7$, and $ BC = 8$, line $ OI$ intersects the smallest and largest sides. This is trivial to check and is left as an exercise to the reader. We now show that we can continuously move $ A$, $ B$, and $ C$ through the plane while keeping the triangle acute and scalene to make $ AB$ and $ AC$ have any positive real lengths $ c < b < 8$. To increase or decrease any side, we rotate one of its vertices about the opposite vertex while keeping the other vertices stationary. This moves one of the vertices along the arc of a circle, which is a continuous motion, and we can achieve any desired length assuming that keeps the triangle non-degenerate. We then proceed by casework to avoid making the triangle isosceles: If $ c > 5$ and $ b > 7$ we increase $ AC$ to $ b$ and then $ AB$ to $ c$. If $ c > 5$ and $ b < 7$ we increase $ AB$ and then decrease $ AC$. If $ c < 5$ and $ b > 7$ we increase $ AC$ and then decrease $ AB$. If $ c < 5$ and $ b < 7$ we decrease $ AB$ and then decrease $ AC$. This will keep $ BC = 8$ as the largest side since $ b,c,5,7 < 8$ and will thus cause the triangle to remain acute since we increase $ AB$ and $ AC$ before doing any decreasing. It also avoids any isosceles triangles in the path, as one can check in each of the case. It thus holds that in any acute scalene triangle with greatest side $ 8$, $ OI$ intersects the smallest and largest side. We can then appropriately scale the triangle to account for all possible scalene triangles, and the result is proved. QED

I don't think that continuous transformation stuff is that rigirous? What do you mean by it exactly? We can also easily use barycentric coordinates to solve this problem.

By continuous transformation I just mean the definition from topology, using standard real topology. I guess by "continuously move" the vertices I should say move them along a connected path. But yeah what we're basically trying to show is that if we move it continuously and change the sides that $ OI$ intersects then $ OI$ would have to intersect one of the vertices, so using the geometric fact I proved at the beginning, we can just prove it for one triangle, show we can move $ A$, $ B$, and $ C$ to any scalene triangle and thus prove it for all of them.

Is there a way to do this without topology? I'm confused.

Edit: Is essentially the same as froufrou4189. Meh, I'm still happy. Edit2: Oops, meant intersection not union.

#H34N1 wrote: Is there a way to do this without topology? I'm confused. Uh, the solutions don't directly use topology, he just called upon a definition to make it more rigorous.

#H34N1 wrote: Is there a way to do this without topology? I'm confused. Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=261335187&t=125578 #H34N1 wrote: Let $ ABC$ be an acute-angled triangle whose side lengths satisfy the inequalities $ AB < AC < BC$. If point $ I$ is the center of the inscribed circle of triangle $ ABC$ and point $ O$ is the center of the circumscribed circle, prove that line $ IO$ intersects segments $ AB$ and $ BC$. PostPosted: Sun Dec 24, 2006 9:45 pm There is a certain amount of irony here

I do not understand pianoforte's solution that I must be within some region for OI to intersect BC. Can someone provide a clear solution(I am horrible at geometry).