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### Triangle and Interior Point

```
Date: 03/23/99
From: Anonymous
Subject: Proof involving a Triangle and an interior Point

Let P be an interior point of a triangle ABC. Lines AP, BP and CP meet
three sides BC, CA and AB at R, S and T, respectively. Does the
following geometric inequality hold?

Area of triangle RST <= 1/4 of Area of Triangle ABC.
The equality holds if and only if P is the center of triangle ABC.

I have examined some special cases and found out that the statement is
true. Could you help me find a general proof or a counter-example?
```

```
Date: 03/23/99
From: Doctor Floor
Subject: Re: Proof involving a Triangle and an interior Point

For the proof we will use two famous results:

1. Ceva's Theorem. This theorem says that if triangles ABC and RST are
as you described (so AR, BS and CT intersect in one point), then

AT   BR   CS
-- * -- * -- = 1.
TB   RC   SA

2. The geometric mean is always smaller than or equal to the
arithmetic mean. For three numbers p, q and r that means:

(pqr)^(1/3) <= (p + q + r)/3
pqr <= [(p + q + r)/3]^3

The equality holds if and only if p=q=r.

Now, let's see how to apply these results:

Let a = BC, b = AC and c = AB. Let p, q and r be given by AT = p*c,
BR = q*a and CS = r*b. We can suppose that p + q + r <= 1.5.
[If this inequality does not hold, then we can let p', q', r' be given
by p' = 1 - p (BT = p'*c), q' = 1 - q (CR = q'*a) and r' = 1 - r
(CS = r*b); now we will find p' + q' + r' <= 1.5 and the rest of the
proof can be done with p', q', r').

From Ceva's theorem we find that:

pqr
--------------------- = 1
(1 - p)(1 - q)(1 - r)

pqr = (1 - p)(1 - q)(1 - r)

pqr = 1 - p - q - r + pq + pr + qr - pqr

p + q + r - pq - pr - qr = 1 - 2pqr  [1].

Now, note that area(TBR) = q(1 - p)*area(ABC). This can be seen
because area(ABR) = q*area(ABC) and area(TBR) = (1 - p)*area(ABR).

In the same way we find:

area(SRC) = r(1 - q)*area(ABC)
area(ATS) = p(1 - r)*area(ABC)

So area(TBR) + area(SRC) + area(ATS)
= (p + q + r - pq - pr - qr)*area(ABC)
= (1 - 2pqr)*area(ABC) (see [1]).

This gives us, in the end, that area(RST) = 2pqr*area(ABC)  [2].

Earlier, we found (p + q + r) <= 1.5. This gives:

(p + q + r)/3  <= 0.5
[(p + q + r)/3]^3 <= 0.125

Applying that the geometric mean is always smaller than or equal to
the arithmetic mean, we find:

pqr <= [(p + q + r)/3]^3 <= 0.125

Inserting this in [2] gives:

area(RST) <= 0.25*area(ABC) [3]

Using that we know that pqr = [(p + q + r)/3]^3 if and only if
p = q = r, we see that the equality in [3] holds if and only if
p = q = r = 0.5 (if p = q = r<0.5, then Ceva's theorem does not hold).

This proves the statement.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 03/24/99
From: Anonymous
Subject: Re: Proof involving a Triangle and an interior Point

Thank you very much for your help. I would like to add a little

Since pqr = (1 - p)(1 - q)(1 - r) and area(RST) = 2pqr*area(ABC), we
have area(RST) = 2*[pqr(1-p)(1-q)(1-r)]^(1/2)area(ABC). Note that
x(1 - x) <= 1/4 for 0 < x < 1, the equality holds if and only if
x = 1/2. Thus, area(RST) <= 1/4*area(ABC). The equality holds if and
only if p = q = r = 1/2, which means the point P is in the center.
```

```
Date: 03/24/99
From: Doctor Floor
Subject: Re: Proof involving a Triangle and an interior Point

This is an excellent and sharp simplification of the last part of the
proof. There is no need any more to assure that (p + q + r)/3 <= 0.5.

Note that you again use that the geometric mean is smaller than or
equal to the arithmetic mean when you say that x(1-x) <= 1/4. Since
the arithmetic mean of x and 1-x equals 1/2, we find
sqrt(x(1 - x)) <= 1/2 and then x(1 - x) <= 1/4.

I like the way the equality follows from your note.

When p = q = r = 0.5, it means that P is the "centroid" or "center of
gravity." "The center" of a triangle is not well defined. In fact,
there are many triangle centers. You can visit a description of a lot
of triangle centers on a site by Professor Clark Kimberling of
the University of Evansville:

http://www.evansville.edu/~ck6/tcenters/

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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