Question

for which of the following conjectures can a counterexample be found? altitudes of any triangle intersect at a point altitudes of right triangles intersect at the right angle altitudes of acute triangles intersect inside the triangles medians and altitudes intersect in a common point

Explanation:

1. For "altitudes of any triangle intersect at a point": The orthocenter (intersection of altitudes) exists for all triangles, so no counterexample.
2. For "altitudes of right triangles intersect at the right angle": In a right triangle, two altitudes are the legs, and the third is from the right angle to the hypotenuse. All three altitudes intersect at the right - angle vertex, so no counterexample.
3. For "altitudes of acute triangles intersect inside the triangles": In an acute triangle, all altitudes are inside the triangle, so no counterexample.
4. For "medians and altitudes intersect in a common point": Consider a non - isosceles triangle, say a scalene triangle. The median (line from vertex to mid - point of opposite side) and the altitude (perpendicular from vertex to opposite side) from the same vertex will be different lines (except in isosceles triangles where they coincide). So, we can find a counterexample here (e.g., take a scalene triangle like a triangle with sides 3, 4, 5 (wait, 3 - 4 - 5 is a right triangle, take a scalene acute triangle like sides 2, 3, 4). The median and altitude from the vertex opposite the side of length 4 will not be the same line, so they don't intersect at the same point (except in isosceles triangles). So this conjecture is false and a counterexample can be found.

Answer:

medians and altitudes intersect in a common point