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# What happens when $$R=h_A$$?(Basics of Construction)

Prerequisite: Triangles - Circumcenter

In this series of notes, I will use pure geometry to deduce a few properties in the configuration where a triangle's circumradius is equal to one of its heights. Our discussion will pertain to the main triangle $$ABC$$ with circumradius $$R$$, circumcenter $$O$$, and the height(altitude) of $$A$$ is denoted $$h_A$$.

To begin, we ask an essential question: when can $$R=h_A$$ occur? Can it happen when $$\angle A$$ is acute? Right? Obtuse?

Just by drawing a simple diagram, we know it cannot happen when $$\angle A$$ is obtuse. If $$\angle A$$ is obtuse, then $$A,O$$ are on opposite sides of $$BC$$ by properties of circumcenters. This means $$AO$$ must be larger than the closest distance from $$A$$ to $$BC$$, i.e $$h_A$$.

When $$\angle A$$ is right, $$O$$ becomes the midpoint of $$BC$$; thus the altitude and median must coincide, which makes $$BAC$$ an isosceles right triangle.

For the case $$\angle A$$ is acute, the truth is that it can always happen. We will prove this below by finding a way to construct $$A$$ given $$BC$$ and $$O$$. The rest of the discussion will assume that $$A$$ is acute. In the meantime, you are welcome to prove it any other way you like.

To kick the pure geometry off, let $$K$$ denote the base of the altitude of $$A$$. Rightaway, we will invoke the internal bisector of $$\angle BAC$$, which also bisects $$\angle KAO$$(property of circumcenter). Let the bisector intersect the circumcircle $$(O)$$ again at $$N$$, then $$ON\perp BC$$ because $$N$$ is the midpoint of minor arc $$BC$$. Hence $$ON\parallel AK, ON=AK$$, which implies $$AONK$$ is a parallelogram; In fact, it is a rombus because $$AO=NO$$. This gives us a crucial property to motivate our construction: $$NO=NK$$.

Construction Problem: Given a circle and on it a fixed chord $$BC$$. Construct $$A$$ such that $$R=h_A$$.

Solution: We know $$A$$ must lie on the major arc $$BC$$. First, construct $$N$$ the midpoint of minor arc $$BC$$. Next, draw the circle centered at $$N$$ with radius $$ON$$ intersect $$BC$$ at two point(because $$O,N$$ are on opposite sides of $$BC$$), denote one of the points $$K$$. Finally, construct perpendicular bisector of $$OK$$ and it intersects the major arc $$BC$$ at $$A$$, our desired construction. This ends our the construction portion.

Now we prove that $$A$$ indeed satisfies $$R=h_A$$. Since $$AO=NO$$, therefore $$OK$$ is the perpendular bisector of $$AN$$. Thus $$AONK$$ is a rombus and $$AK\parallel ON$$, which means $$AK$$ is the altitude of $$A$$ and $$R=ON=AK=h_A$$ $$._{\Box}$$

In conclusion, the provided construction should make it clear that $$\angle A$$ is acute or right is a necessary condition for $$R=h_A$$.

In the next note of this series, we will add the incenter of $$\triangle ABC$$ to the picture and derive a few awesome properties. In the mean time, ponder this:

Suppose $$R=h_A$$ and $$I$$ is the incenter of $$ABC$$, let $$D$$ be the foot of projection of $$I$$ on $$BC$$. Prove that $$OD\perp ND$$, where $$N$$ is still the midpoint of arc $$BC$$.

Note by Xuming Liang
1 year ago

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Great geometrical construction! @Xuming Liang.Seeing this I thought of constructing a triangle with $$m_{a}=R$$($$m_{a}$$ is median from vertex A),which I almost got,but I could construct only obtuse and right angled triangles,but not for the case of acute angled triangle.Is it possible to construct it?Can you help me!:D · 1 year ago

$$m_a=R$$ is interesting as well. To get an idea for the construction, we label the midpoint of $$BC$$ as $$M$$ and the circumcenter of $$ABC$$ as $$O$$; thus the condition is $$AO=AM$$, which means $$A$$ lies on the perpendicular bisector of $$OM$$. How do you use this new information about $$A$$ to help with the construction? · 1 year ago

I used exactly similar approach,by constructing the perpendicular bisector of the perpendicular dropped on chord $$BC$$ from the circumcenter and then joining $$B,C$$ to $$A$$,that is where the perpendicular bisector of $$OM$$ meets the circumference. I think this point$$A$$ is only possible to get in the semi-circumference containing chord $$BC$$,which gives an obtuse angled triangle unless $$BC$$ is not the diameter.

If $$BC$$ is the diameter then any point $$A$$ on the circumference satisfies $$m_{a}=R$$:) · 1 year ago