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 \(BC\).

## Comments

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TopNewestGreat 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 – Siddharth Singh · 8 months, 3 weeks ago

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– Xuming Liang · 8 months, 3 weeks 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?Log in to reply

If \(BC\) is the diameter then any point \(A\) on the circumference satisfies \(m_{a}=R\):) – Siddharth Singh · 8 months, 3 weeks ago

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