Geometric Inequalities in Pedal Quadrilaterals

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Forum Geometricorum Volume 18 (2018) 71–82. b

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FORUM GEOM ISSN 1534-1178

Geometric Inequalities in Pedal Quadrilaterals S¸ahlar Meherrem, Gizem Günel Açıksöz, Serenay S¸en, Zeynep Sezer, and Günes¸ Bas¸kes

Abstract. The aim of this paper is to investigate the general properties of the pedal quadrilateral of a point P with respect to a convex and closed quadrilateral ABCD. In particular, we present an analogue of Erd˝os-Mordell Inequality stating that for any triangle ABC and a point P inside ABC, the sum of the distances from P to the sides is less than or equal to half of the sum of the distances from P to the vertices of ABC, for an inscribed quadrilateral.

1. Introduction Let ABC be a triangle and P be a point in plane. If the feet of the perpendiculars drawn from P to sides of the triangle [BC], [CA], [AB] are respectively X, Y and Z, then triangle XY Z is called the pedal triangle of P with respect to ABC and P is called the pedal point ([3, 7]). If the point P lies on the circumcircle of the triangle ABC, then the points X, Y and Z are collinear. The line passing through the points X, Y and Z is known as the Simson line of P ([3]). Z A

A

P

Y

Z

B

Y P

B X

C

X

C

Figure 1.

A similar notion has been generalized for polygons with n sides ([3]). Let A1 A2 A3 · · · An be a polygon and P be a point inside A1 A2 A3 · · · An . If the feet of the perpendiculars from P to the line segments [A1 A2 ], [A2 A3 ], . . . , [An A1 ] are respectively H1 , H2 , . . . , Hn , then the polygon H1 H2 · · · Hn is called the pedal polygon of P with respect to A1 A2 A3 · · · An and P is called the pedal point ([3]). Although there are lots of investigations on pedal triangles, so little investigations have been studied on pedal quadrilaterals in the literature ([3, 5, 6, 7, 9, 11]). Publication Date: February xx, 2018. Communicating Editor: Paul Yiu.

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In this paper, we study on the general properties of the pedal quadrilateral of a point P with respect to a convex and closed quadrilateral ABCD. We investigate some interesting geometric inequalities on pedal quadrilaterals. In particular, we state an analogue of Erd˝os-Mordell Inequality ([1, 2, 4, 8, 10]) for quadrilaterals. 2. Geometric inequalities in pedal quadrilaterals Proposition 1. Let ABCD be an inscribed quadrilateral and P be a point inside ABCD. If the feet of perpendiculars from P to the line segments [AB], [BC], [CD] and [AD] are respectively K, M , N , L and the radius of the circumcircle of ABCD is R then, |KM | =

|AC|·|P B| , 2R

|KL| =

|BD|·|P A| , 2R

|M N | =

|BD|·|P C| , 2R

|LN | =

|AC|·|P D| . 2R

B K

M

P

A

C L N

D

Figure 2.

Proof. As seen in Figure 2 and since ∠BKP = ∠P LD = ∠P N C = ∠P M B = 90◦ , the quadrilaterals BKP M , P LDN , P N CM and P LAK are inscribed quadrilaterals. By the law of sines, we have |P B| =

KM sin KP M

and

2R =

Hence, we get

|AC| |AC| = . sin(π − ∠KP M ) sin KP M

|KM | =

|AC| · |P B| . 2R

|M N | =

|BD| · |P C| , 2R

It can be easily seen that |KL| =

|BD| · |P A| , 2R

|LN | =

|AC| · |P D| . 2R

Geometric inequalities in pedal quadrilaterals

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Proposition 2. Let ABCD be a quadrilateral and P be a point inside ABCD.If the feet of perpendiculars from P to the line segments [AB], [BC], [CD] and [AD] are respectively A1 , B1 , C1 , D1 , then M ·C r1 + r2 + r3 + r4 ≤ √ , 2 2 where r1 , r2 , r3 and r4 are respectively the radii of the incircles of the triangles AA1 D1 , A1 BB1 , CB1 C1 , DC1 D1 , and M = max

1 1 1 1 √ , √ √ , √ √ , √ √ √ 1 − cos A + 2 1 − cos B + 2 1 − cos C + 2 1 − cos D + 2

and C is the perimeter of the quadrilateral ABCD. The equality holds if ABCD is chosen as a square and P is chosen at the center of the square. B O2 A1 r2 B1 A O1

r1 O3 r3

P

C

D1 r4

C1

O4 D

Figure 3.

Proof. By the law of cosines, we have p |A1 B1 | = |A1 B|2 + |BB1 |2 − 2|A1 B||BB1 | cos B.

(1)

If SA1 BB1 is the area of the triangle A1 BB1 , then

|A1 B| · |BB1 | sin B . 2 Since 2SA1 BB1 = (|A1 B| + |BB1 | + |A1 B1 |)r2 , we get 2SA1 BB1 . r2 = |A1 B| + |BB1 | + |A1 B1 | By the equalities (2) and (3), |A1 B| · |BB1 | sin B r2 = . |A1 B| + |BB1 | + |A1 B1 | It follows from Arithmetic-Geometric Mean Inequality that SA1 BB1 =

|A1 B|2 + |BB1 |2 ≥ 2 · |A1 B| · |BB1 |.

(2)

(3)

(4)

(5)

,

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If the inequalities (1) and (5) are considered together, then p |A1 B1 | = |A1 B|2 + |BB1 |2 − 2|A1 B| · |BB1 | cos B p 2|A1 B| · |BB1 | − 2|A1 B| · |BB1 | cos B ≥ √ p √ ≥ 2 |A1 B| · |BB1 | 1 − cosB

It is also obvious from Arithmetic-Geometric Mean Inequality that p |A1 B| + |BB1 | ≥ 2 |A1 B| · |BB1 |.

(6)

(7)

Therefore, by the inequalities (6) and (7), we observe that p √ p √ |A1 B|+|BB1 |+|A1 B1 | ≥ 2 |A1 B| · |BB1 |+ 2 |A1 B| · |BB1 | 1 − cos B. (8) Hence, it follows from the inequalities (4) and (8) that sin B p p √ 2 · |A1 B| · |BB1 | 1 − cos B + 2 |A1 B| · |BB1 | sin B √ ≤ |A1 B| · |BB1 | · √ p √ 2 |A1 B| · |BB1 | 1 − cos B + 2) p sin B √ ≤ |A1 B| · |BB1 | · √ √ 2( 1 − cos B + 2) |A1 B| + |BB1 | √ ≤ √ √ (9) 2 2( 1 − cos B + 2)

r2 ≤ |A1 B| · |BB1 | √

Let

M = max

By inequality (9),

√

1 1 √ , √ √ , 1 − cos A + 2 1 − cos B + 2 1 1 √ , √ √ √ . 1 − cos C + 2 1 − cos D + 2 r2 ≤

M · (|A1 B| + |BB1 |) √ . 2 2

Similarly, we get r1 ≤

M ·(|A1 A|+|AD1 |) √ , 2 2

r3 ≤

M ·(|C1 C|+|CB1 |) √ , 2 2

r4 ≤

M ·(|C1 D|+|DD1 |) √ . 2 2

              

If the inequalities (10) and (11) are added side by side, it is easy to see that M ·C √ 2 2 where C is the perimeter of the quadrilateral ABCD. r1 + r2 + r3 + r4 ≤

(10)

(11)


a

B

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a

B1

C

O2

O3

r2 r3

a

A1

a

P C1

a

a

r1 r4 O1

A

O4

a

D1

a

D

Figure 4.

Now suppose that ABCD is a square with side length 2a and P is at the center of the ABCD (Figure 4). If the area of the triangle D1 AA1 is SD1 AA1 then, √ 2a + a 2 a2 SD1 AA1 = r1 = . 2 2 1 a√ , Hence, we get r1 = 2+ 2 . Since r1 = r2 = r3 = r4 , C = 8a and M = √2+1 then the following equality holds: √1 8a · a C ·M 2+1 √ = √ r1 + r2 + r3 + r4 = 4 · = √ . 2+ 2 2 2 2 2 Proposition 3. Let A1 A2 A3 ·An be a polygon and P be a point inside A1 A2 A3 · · · An . If the feet of perpendiculars from P to the line segments [A1 A2 ], [A2 A3 ], . . . , [An A1 ] are respectively H1 , H2 , . . . , Hn , then M ·C √ 2 2 where r1 , r2 , . . . , rn are respectively the radii of the incircles of the triangles A1 H1 Hn , H1 A2 H2 , . . . , Hn−1 An Hn , and 1 1 1 √ , √ √ , ..., √ √ M = max √ 1 − cos A1 + 2 1 − cos A2 + 2 1 − cos An + 2 and C is the perimeter of the polygon A1 A2 · · · An . The equality holds if A1 A2 · · · An is chosen as a square and P is chosen as the centroid of the square. r1 + r2 + · · · + rn ≤

This can be easily shown as in Proposition 2.

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Proposition 4. Let ABCD be a circumscribed quadrilateral with area S. If the center and the radius of the incircle of ABCD are respectively the point P and r, and the feet of perpendiculars from P to the line segments [AB], [BC], [CD] and [AD] are respectively K, L, M and N , then Spedal 1p 2 ≤1− a + b2 + c2 + d2 , S 4r where a = |BL|, b = |LC|, c = |M D|, d = |AN |, and Spedal is the area of the pedal quadrilateral KLM N . The equality holds if ABCD is a square. B

a L

b C

a S1

S2

K

b P

S4

d

M

S3 c

A

d

N

c

D

Figure 5.

Proof. As seen in Figure 5, if S1 , S2 , S3 and S4 are respectively the areas of triangles KP L, LP M , M P N and N P K, then we get a2 sin B , 2 b2 sin C , S2 = b · r − 2 c2 sin D S3 = c · r − , 2 d2 sin A . S4 = d · r − 2 S1 = a · r −

Hence, it is obvious that 1 S1 +S2 +S3 +S4 = r(a+b+c+d)− (a2 sin B +b2 sin C +c2 sin D +d2 sin A). 2 Since we also have S = (a + b + c + d)r, we get Spedal S1 + S2 + S3 + S4 = S S 2 a sin B + b2 sin C + c2 sin D + d2 sin A = 1− . (12) 2r(a + b + c + d)


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By the Cauchy-Schwarz Inequality, we get

Thus,

a+b+c+d= a·1+b·1+c·1+d·1 p p a2 + b2 + c2 + d2 · 12 + 12 + 12 + 12 ≤ p = 2 a 2 + b2 + c 2 + d 2 . −

1 1 . ≤− √ 2 2 a+b+c+d 2 a + b + c 2 + d2

By the equality (12), Spedal a2 sin B + b2 sin C + c2 sin D + d2 sin A = 1− S 2r(a + b + c + d) 2 2 a + b + c 2 + d2 ≤1− √ 4r a2 + b2 + c2 + d2 1p 2 a + b2 + c2 + d2 . = 1− 4r B

a

a

a

L

r

C

a

P

r

K

M

r a

A

r

a

N

a

a

D

Figure 6.

Now suppose that ABCD is a square and |AB| = |BC| = |CD| = |AD| = 2a. As seen in Figure 6, since r = a, S = 4a2 , Spedal = 2a2 , we have Spedal 1p 2 2a2 1 1p 2 a + a2 + a2 + a2 = 1− a + b2 + c2 + d2 . = 2 = = 1− S 4a 2 4a 4r Proposition 5. Let P be a point inside a polygon A1 A2 A3 · · · An where |A1 A2 | = a1 , |A2 A3 | = a2 , . . . , |An A1 | = an . If the lengths of the perpendiculars from P to the sides of the polygon A1 A2 A3 · · · An are respectively h1 , h2 , . . . , hn , then S 1 1 ≥ · a1 2 C 2 h1 + · · · + hann

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where S and C are respectively the area and the perimeter of A1 A2 A3 · · · An . The equality holds if a1 = a2 = · · · = an = a and h1 = h2 = · · · = hn = h. A3

a2 a3

A2

A4

a1

h2 h1

h3 a4

A1

h4

P

A5

Figure 7.

Proof. As seen in Figure 7, it is obvious that 2S = a1 h1 + a2 h2 + · · · + an hn . By the Cauchy-Schwarz Inequality, we get q q p a1 a2 an (a1 h1 +a2 h2 +· · ·+an hn ) ≥ ( a21 + a22 +· · ·+ a2n )2 . + + ··· + h1 h2 hn Since 2S

a2 an a1 + + ··· + h1 h2 hn

we have 2S ≥ Hence,

≥(

q

a21 +

(a1 + a2 + · · · + an )2 a1 a2 an = h1 + h 2 + · · · + h n 1 S ≥ · 2 C 2

a1 h1

+

a2 h2

a1 h1

q p a22 + · · · + a2n )2

+

1 + ··· +

a2 h2

C2 + ··· +

an hn

an hn

.

.

Now suppose that a1 = a2 = · · · = an = a, and h1 = h2 = · · · = hn = h. Since, ah , C = na, S =n· 2 we get S 1 nah h 1 1 1 = · a1 . = 2 2 = = · a a C2 2n a 2na 2 h + h + · · · + ha 2 h1 + ha22 + · · · + hann


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Proposition 6. Let P be a point inside a polygon A1 A2 A3 · · · An where |A1 A2 | = a1 , |A2 A3 | = a2 , . . . , |An A1 | = an . If the lengths of the perpendiculars from P to the sides of the polygon A1 A2 A3 · · · An are respectively h1 , h2 , . . . , hn , then 1 1 n2 1 + + ··· + ≥ a1 h1 a2 h2 an hn 2S where S is the area of A1 A2 A3 · · · An . The equality holds if a1 = a2 = · · · = an = a and h1 = h2 = · · · = hn = h. Proof. By the Cauchy-Schwarz Inequality, we have 1 1 1 2 ≥ (1 + + ··· + (a1 h1 +a2 h2 +· · ·+an hn ) {z· · · + 1}) . |+1+ a1 h1 a2 h2 an hn n times

Since 2S = a1 h1 + a2 h2 + · · · + an hn , it is easy to see that

1 1 1 n2 + + ··· + ≥ . a1 h1 a2 h2 an hn 2S Now suppose that a1 = a2 = · · · = an = a and h1 = h2 = · · · = hn = h. Since S = n · ah 2 , we have 1 1 1 1 n2 n2 + + ··· + =n· = = . a1 h1 a2 h2 an hn ah nah 2S

Theorem 7. Let ABCD be an inscribed quadrilateral and P be a point inside ABCD. If the feet of perpendiculars from P to the line segments [AB], [BC], [CD] and [AD] are A1 , B1 , C1 , D1 respectively, then p |P A| + |P B| + |P C| + |P D| > 4 4 |P A1 | · |P A2 | · |P A3 | · |P A4 |. B

A2

A1

A P

C

A4 A3 D

Figure 8.

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Proof. Let |P A1 | = h1 , |P A2 | = h2 , |P A3 | = h3 and |P A4 | = h4 . Then by the law of cosines, we have q h21 + h22 − 2h1 h2 cos(π − B), q h22 + h23 − 2h2 h3 cos(π − C), |A2 A3 | = q |A3 A4 | = h23 + h24 − 2h3 h4 cos(π − D), q |A4 A1 | = h24 + h21 − 2h1 h4 cos(π − A). |A1 A2 | =

Hence, |A1 A2 |2 = h21 + h22 − 2h1 h2 cos(π − B)

= h21 + h22 − 2h1 h2 cos(A + C + D − π)

= h21 + h22 + 2h1 h2 cos(A + C + D)

= h21 + h22 + 2h1 h2 cos(A + C) cos D − 2h1 h2 sin(A + C) sin D = (h2 sin D − h1 sin(A + C))2 + (h2 cos D + h1 cos(A + C))2 .

Since A + C = π, we have |A1 A2 |2 = (h2 sin D)2 + (h2 cos D − h1 )2 . Now suppose that (h2 cos D − h1 )2 = 0. Then, cos D = hh12 . Since ∠A1 P A2 = ∠D, by the law of cosines, |A1 A2 |2 = h21 + h22 − 2h1 h2 cos D h1 = h21 + h22 − 2h1 h2 · h2 2 2 = h2 − h1 . Hence, we get h22 = |A1 A2 |2 + h21 , implying that ∠P A1 A2 = 90◦ , which is a contradiction. Therefore, (h2 cos D − h1 )2 6= 0. Then, it is easy to verify that |A1 A2 | > h2 sin D

(13)

|A2 A3 | > h3 sin A

(14)

|A1 A4 | > h1 sin C

(16)

Similarly, we have

|A3 A4 | > h4 sin B

(15)

Since P A1 BA2 , P A2 CA3 , P A3 DA4 and P A4 AA1 are inscribed quadrilaterals, if we apply the law of sines to the triangles AA1 A4 , BA1 A2 , CA2 A3 , DA3 A4 ,


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then we respectively get, |A1 A4 | (17) sin A |A1 A2 | |P B| = (18) sin B |A2 A3 | (19) |P C| = sin C |A3 A4 | |P D| = (20) sin D If the inequalities (13), (14), (15) and (16) are respectively considered together with the inequalities (17), (18), (19) and (20), then |P A| =

|A1 A2 | sin B |A2 A3 | |P C| = sin C |A3 A4 | |P D| = sin D |A1 A4 | |P A| = sin A Therefore, it is easy to see that |P B| =

h2 sin D , sin B h3 sin A > , sin C h4 sin B > , sin D h1 sin C > . sin A >

h1 sin C h2 sin D h3 sin A h4 sin B + + + sin A sin B sin C sin D r h sin D h sin A h h sin C 4 2 3 4 sin B 1 · · · > 4 sin A sin B sin C sin D p 4 = 4 |P A1 | · |P A2 | · |P A3 | · |P A4 |

|P A| + |P B| + |P C| + |P D| >

by Arithmetic-Geometric Mean Inequality.

References [1] C. Alsina and R. B. Nelsen, A visual proof of the Erd˝os-Mordell inequality, Forum Geom., 7 (2007) 99–102. [2] L. Bankoff, An elementary proof of the Erd˝os-Mordell theorem, Amer. Math. Monthly, 65 (1958), 521. [3] H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited, Math. Assoc. America, 1967. [4] P. Erd˝os, Problem 3740, Amer. Math. Monthly, 42 (1935) 396. [5] D. Ferrarello, M. F. Mammana, and M. Pennisi, Pedal poygons, Forum Geom., 13 (2013) 153– 164. [6] R. Honsberger, Episodes In Nineteenth and Twentieth Century Euclidean Geometry, Math. Assoc. America, 1995. [7] R. A. Johnson, Advanced Euclidean Geometry, Dover Publications, 2007. [8] D. K. Kazarinoff, A simple proof of the Erd˝os-Mordell inequality for triangles, Michigan Math. Journal, 4 (1957) 97–98. [9] M. F. Mammana, B. Micale, and M. Pennisi, Orthic quadrilaterals of a convex quadrilateral, Forum Geom., 10 (2010) 79–91.

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[10] L. J. Mordell and D. F. Barrow, (1937). Solution to Problem 3740, Amer. Math. Monthly, 44 (1937) 252–254. [11] M. S¸ahin, Matematik Olimpiyatlarına Hazırlık Geometri - I, Palme Yayınları, Ankara, 2000. S¸ahlar Meherrem: Yasar University, Faculty of Science and Letters, Department of Mathematics, Izmir, Turkey E-mail address: [email protected] Gizem Günel Açıksöz: Ozel Ege Lisesi, Izmir, Turkey E-mail address: [email protected] Serenay S¸en : Ozel Ege Lisesi, Izmir, Turkey E-mail address: [email protected] Zeynep Sezer : Ozel Ege Lisesi, Izmir, Turkey E-mail address: gunes [email protected] Günes¸ Bas¸kes: Ozel Ege Lisesi, Izmir, Turkey E-mail address: [email protected]