Triangle and Interior PointDate: 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 note to your proof. 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/ |
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