What happens when R=hAR=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 ABCABC with circumradius RR, circumcenter OO, and the height(altitude) of AA is denoted hAh_A.

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

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

When A\angle A is right, OO becomes the midpoint of BCBC; thus the altitude and median must coincide, which makes BACBAC an isosceles right triangle.

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

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

Construction Problem: Given a circle and on it a fixed chord BCBC. Construct AA such that R=hAR=h_A.

Solution: We know AA must lie on the major arc BCBC. First, construct NN the midpoint of minor arc BCBC. Next, draw the circle centered at NN with radius ONON intersect BCBC at two point(because O,NO,N are on opposite sides of BCBC), denote one of the points KK. Finally, construct perpendicular bisector of OKOK and it intersects the major arc BCBC at AA, our desired construction. This ends our the construction portion.

Now we prove that AA indeed satisfies R=hAR=h_A. Since AO=NOAO=NO, therefore OKOK is the perpendular bisector of ANAN. Thus AONKAONK is a rombus and AKONAK\parallel ON, which means AKAK is the altitude of AA and R=ON=AK=hAR=ON=AK=h_A .._{\Box}

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

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

Suppose R=hAR=h_A and II is the incenter of ABCABC, let DD be the foot of projection of II on BCBC. Prove that ODNDOD\perp ND, where NN is still the midpoint of arc BCBC.

Note by Xuming Liang
5 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

Great geometrical construction! @Xuming Liang.Seeing this I thought of constructing a triangle with ma=Rm_{a}=R(mam_{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 - 5 years, 6 months ago

Log in to reply

ma=Rm_a=R is interesting as well. To get an idea for the construction, we label the midpoint of BCBC as MM and the circumcenter of ABCABC as OO; thus the condition is AO=AMAO=AM, which means AA lies on the perpendicular bisector of OMOM. How do you use this new information about AA to help with the construction?

Xuming Liang - 5 years, 6 months ago

Log in to reply

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

If BCBC is the diameter then any point AA on the circumference satisfies ma=Rm_{a}=R:)

Siddharth Singh - 5 years, 6 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...