An extension of Carnot’s theorem Tran Quang Hung Student in department of Mathematics University of Science Ha Noi national university January 28, 2005

Abstract We consider the following problem and the special cases of it as the applications. Let I be an arbitrary point in plane of triangle ABC such that I(α, β, γ) in barycentric coordinates, let da , db , dc be signed distance from I to side BC, CA, AB xαbc + yβca + zγab of triangle ABC. Then xda + ydb + zdc = with circumradii 2R(α + β + γ) R.

1

Introduction

There is a nice indentiy of triangle that we usually call as Carnot’s theorem. In this article we will present an exntension of this indentiy and some apllications of it. Theorem 1. (Carnot’s theorem) Let O be circumcenter of an acute triangle ABC and da , db , dc are distances from O to side BC, CA, AB then da + db + dc = R + r with circumradii R and inradii r of traingle ABC. If triangle is acute then O is inside it, and we consider da , db , dc > 0. In general, we can change circumcenter O by an arbitrary point I in plane of ABC, consider circumcenter O as O(sin 2A, sin 2B, sin 2C) in barycentric coordinates, hence let I(α, β, γ) and furthermore using concept of singed distance with an arbitrary point to sidelines of triangle ABC. With a line d and distances from two points to d, we called they have the same sign if they are in a same halfplane (of d) and have the opposite sign if they are in two different halfplane (of d). With the sidelines of ABC we always consider the distance in halfplane that contained triangle has positive sign.

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An extension of Carnot’s theorem - Tran Quang Hung

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Theorem 2. Let I be an arbitrary point inside triangle ABC such that I(α, β, γ) in barycentric coordinates, let da , db , dc be distances from I to side BC, CA, AB of triangle ABC. Then xαbc + yβca + zγab xda + ydb + zdc = 2R(α + β + γ) with circumradii R And an extension of Carnot’s theorem with signed distance. Theorem 3. Let O be circumcenter of triangle ABC and da , db , dc are distances from O to side BC, CA, AB (da , db , dc ∈ R) then da + db + dc = R + r with circumradii R and inradii r of traingle ABC.

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Proofs

In this section we give the proofs of theorems we presented . Proof.(Carnot’s thereom) A

B'

C' dc

O

db da

B

A'

C

Figure 1: Carnot’s theorem. Let A0 , B 0 , C 0 be midpoint of BC, CA, AB but O is circumcenter of ABC therefore OA0 ⊥ BC, OB 0 ⊥ CA, OC 0 ⊥ AB. Thus we easily seen the quadrilaterals AB 0 OC 0 , BC 0 OA0 , CA0 OB 0 are cyclic. Now apply Ptoleme thereom for them we have c b a a c b b a c db + dc = R , dc + da = R , da + db = R 2 2 2 2 2 2 2 2 2 Now add them we get da

b+c c+a a+b a+b+c + db + dc =R 2 2 2 2

An extension of Carnot’s theorem - Tran Quang Hung

⇒ (da + db + dc )

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ada + bdb + cdc a+b+c = + Rp = S + Rp = p(R + r) 2 2 ⇒ da + db + dc = R + r

. Proof.(Theorem 2) We will use concept of signed area. It issimilar to signed distance, with a segment XY and two arbitrary points P, Q we call area of triangles P XY and QXY have the same sign if P, Q are in the same halfplane (of d) and have the opposite sign if P, Q are in the different halfplane (of d). There is a relation between signed area and signed distance as following [P XY ] d(P, XY ) = d(Q, XY ) [QXY ] here we denote d(P, XY ), [P XY ] as the signed distance and signed area. A

ha

B

da

C

I

Figure 2: Theorem 2 the extension. Now let ha , hb , hc be altitudes of triangle ABC or we can consider them as signed distances from A, B, C to sideline but obvious they are always positive. With I(α, β, γ) and signed da , db , dc from I to sideline we have da [IBC] α = = ha [ABC] α+β+γ but ha =

bc with circumradii R thus we get 2R da =

αbc 2R(α + β + γ)

An extension of Carnot’s theorem - Tran Quang Hung

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similarly we have db =

βca γab , dc = 2R(α + β + γ) 2R(α + β + γ)

multiply them with the real numbers x, y, z, add them we get the indentiy xda + ydb + zdc =

xαbc + yβca + zγab 2R(α + β + γ)

We are done, the solution is simple but the indenity have many applications. Proof.(Theorem 3) We easily prove it by Theorem 2, with arbitrary triangle then O(sin 2A, sin 2B, sin 2C) in barycentric coordinate, choose x = y = z = 1, let I ≡ O then sin 2Abc + sin 2Bca + sin 2Cab da + db + dc = . 2R(sin 2A + sin 2B + sin 2C) Note that a = 2R sin A, sin 2A = 2 sin A cos A e.t.c and X

sin 2A = 4

Y

sin A,

X

cos A =

R+r R

therefore da + db + dc =

8R2

Q

P X sin A( cos A) Q = R( cos A) = R + r 8R · 4 sin A

Thus this extension of Carnot’s theorem is a very special case of our theorem (theorem 2).

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Some applications

We can see the theorem 3 is an application of our main theorem. In this section we will show some more applications of it in geometric indentities and geometric inequalities. Problem 1. Given triangle ABC, let ga , gb , gc be distance from centroid to sideline of ABC prove that ga gb gc + + ≥ sin A + sin B + sin C. (a) ga + gb + gc ≤ 3r and a b c (b) Forall x, y, z > 0 then r √ R(x + y + z)2 6ga gb gc yzga + zxgb + xygc ≤ ⇒ xga + ygb + zgc ≥ xy + yz + zx . 6 R (c)

1 1 1 6 + + ≥ . ga gb gc R

Proof. (a) Let I ≡ G(1, 1, 1), x = y = z = 1 we have ga + gb + gc =

ab + bc + ca 6R

An extension of Carnot’s theorem - Tran Quang Hung

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A

gc

G

gb ga C

B

Figure 3: With centroid. but we easily prove ab + bc + ca ≥ 18Rr (It is equivalent (a + b + c)(ab + bc + ca) ≥ 27abc) ⇒ ga + gb + gc ≥ 3r. And let 1 1 1 x = , y = , z = ⇒ ga + gb + gc = a b c

bc a

ca b

+

+

ab c

6R

note that b2 c2 + c2 a2 + a2 b2 ≥ 3abc(a + b + c) ⇒

X bc a

X ≥ 3(a + b + c) = 6R( sin A)

we are done. (b) We consider Theorem 2 with I ≡ L(a2 , b2 , c2 ) and x, y, z ∈ R then easily seen xga + ygb + zgc =

xbc + yca + zab . 6R

Now we replace x, y, z by yz, zx, xy then yzga + zxgb + xygc =

yzbc + zxca + xyab . 6R

But X X yzbc = 4R2 ( yx sin B sin C) ≤ R2 (x+y+z)2 ⇒ 6(yzga +zxbgb +xycgc ) ≤ R(x+y+z)2 and now we replace (x, y, z) by (xga , ygb , zgc ) we have R(xga +ygb +zgc )2 ≥ 6(xy+yz +zx)ga gb gc ⇒ xga +ygb +zgc ≥

√

r xy + yz + zx

(c) From result (b) let x = gb , y = gc , z = ga we get ga gb + gb gc + gc ga ≥

6ga gb gc 1 1 1 6 ⇒ + + ≥ . R ga glb gc R

6ga gb gc . R

An extension of Carnot’s theorem - Tran Quang Hung

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Problem 2. Let L be Lemoine point of triangle ABC and la , lb , lc are distance from L to the sideline of ABC prove that (a) la + lb + lc ≤ 3r with inradii r. (b) Forall x, y, z > 0 then r √ 6la lb lc 6(yzla + zxblb + xyclc ) ≤ R(x + y + z)2 ⇒ xla + ylb + zlc ≥ xy + yz + zx R with circumradii R. 1 1 1 6 (c) + + ≥ . la lb lc R

A

L lc

lb la C

B

Figure 4: With Lemoine point. Proof. (a) We consider Theorem 2 with I ≡ L(a2 , b2 , c2 ) and x = y = z = 1 we have la + lb + lc =

abc(a + b + c) 6S a+b+c ≤ = 2S 2 = 3r. 2 2 2 2 2 2R(a + b + c ) a +b +c a+b+c

(b) We consider Theorem 2 with I ≡ L(a2 , b2 , c2 ) (Lemoine point) and x, y, z ∈ R then easily seen xa + yb + zc . xla + ylb + zlc = 2S 2 a + b2 + c 2 Now we replace x, y, z by yz, zx, xy get yzla + zxlb + xylc = 2S But

yza + zxb + xyc (1). a2 + b 2 + c 2

√ a2 + b2 + c2 ≥ 4 3S (2)

and X (x + y + z)2 R √ yza + zxb + xyc = 2R( yz sin A) ≤ 2R = √ (x + y + z)2 (3). 2 3 3

An extension of Carnot’s theorem - Tran Quang Hung

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Combine (1), (2), (3) we get 6(yzla + zxblb + xyclc ) ≤ R(x + y + z)2 and replace (x, y, z) by (xla , ylb , zlc ) we have R(xla + ylb + zlc )2 ≥ 6(xy + yz + zx)la lb lc ⇒ xla + ylb + zlc ≥

√

r xy + yz + zx

6la lb lc . R

(c) From result (b) we let x = lb , y = lc , z = la we get la lb + lb lc + lc la ≥

1 1 6 6la lb lc 1 ⇒ + + ≥ . R la lb lc R

Note that by others choice of x, y, z we can see many other applications of result (b). Problem 3. Let H be othorcenter of triangle ABC and ha , hb , hc are signed distance from H to sideline of ABC prove that (a) Forall x, y, z ∈ R then xha + yhb + zhc = 2R(x cos B cos C + y cos C cos A + z cos A cos B). (b) Forall x, y, z > 0 then ha + hb + hb ≤

2(R + r)2 3R

and y

X cos B cos C ha hb hc 1 yz zx xy +z +x ≤ ( + + ) ⇒ ha + hb + hc ≤ . cos C cos A cos B 2 x y z cos A

(c)

hb hc ha hb hc ha + + = + + = 2(R + r). cos B cos C cos A cos C cos A cos B

Proof. (a) Using Theorem 2 with H(tan A, tan B, tan C) we get P x tan B tan Cbc P xha + yhb + zhc = . 2R( tan A) Note that a = 2R sin A,

X

tan A =

Y

Q sin A tan A = Q cos A

we easily seen xha + yhb + zhc = 2R(x cos B cos C + y cos C cos A + z cos A cos B). (b) From (a) let x = y = z = 1 and X X (R + r)2 3( cos B cos C) ≤ ( cos A)2 = R2

An extension of Carnot’s theorem - Tran Quang Hung

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H

hb hc A ha C

B

Figure 5: With othorcenter. we easily get ha + hb + hb ≤

2(R + r)2 . 3R

Now from (a) we replace (x, y, z) by (

y z x , , ) cos C cos A cos B

and using the fundamental inequality in triangle, forall x, y, z > 0 then X yz zx xy 2( x cos A) ≤ + + x y z we get y

ha hb hc 1 yz zx xy +z +x ≤ ( + + ) cos C cos A cos B 2 x y z

and choose x = cos B, y = cos C, z = cos A we have ha + hb + hc ≤

X cos B cos C cos A

.

(c) From (a) we let x=

1 1 1 ,y = ,z = cos B cos C cos A

x=

1 1 1 ,y = ,z = cos C cos A cos B

and similarly let

and cos A + cos B + cos A = we easily get our result.

R+r R

An extension of Carnot’s theorem - Tran Quang Hung

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Problem 4. Let O be circumcenter of triangle ABC and da , db , dc are signed distance from O to sideline of ABC prove that (a) Forall x, y, z ∈ R then xda + ydb + zdc = R(x cos A + y cos B + z cos C). (b) Forall x, y, z > 0 then xda + ydb + zdc ≤

R yz zx xy yz zx xy ( + + ) ⇒ R( d2a + d2b + d2c ) ≥ 2da db dc (x + y + z). 2 x y z x y z

(c) 2(db dc + dc da + da db ) ≤ R(

db dc dc da da db + + ). da db dc

A

db B

dc

C

da O

Figure 6: With circumcenter. Proof. (a) Using Theorem 2 with I ≡ O(sin 2A, sin 2B, sin 2C) and forall x, y, z ∈ R we have P x sin 2Abc P xd + ydb + zdc = 2R( sin 2A) note that a = 2R sin A, sin 2A = 2 sin A cos A,

X

sin 2A = 4

Y

sin A

we easily get xda + ydb + zdc = R(x cos A + y cos B + z cos C). (b) From (a) and using fundamental inequality X

1 yz zx xy x cos A ≤ ( + + ) 2 x y z

forall x, y, z > 0 we prove it easily, replace (x, y, z) by (xdb dc , ydc da , zda db )

An extension of Carnot’s theorem - Tran Quang Hung we get R(

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yz 2 zx 2 xy 2 d + db + dc ) ≥ 2da db dc (x + y + z) x a y z

which is our result. (c) From (b) let x = db , y = dc , z = da we wil have this result.

References [1] Carnot’s Theorem, http://mathworld.wolfram.com/CarnotsTheorem.html. [2] Nikolaos Dergiades, Signed Distances and the Erdos-Mordell Inequality, ForumGeometricorum, (2004), 66-68. [3] Dragoslav S. Mitrinovic ,J. Pecaric,V. Volenec, Recent Advances in Geometric Inequalities