Pythagoras Theorem and Its Applications

Transcription

1 Lecture 10 Pythagoras Theorem and Its pplications Theorem I (Pythagoras Theorem) or a right-angled triangle with two legs a, b and hypotenuse c, the sum of squares of legs is equal to the square of its hypotenuse, ie a 2 + b 2 = c 2 Theorem II (Inverse Theorem) If the lengths a, b, c of three sides of a triangle have the relation a 2 + b 2 = c 2, then the triangle must be a right-angled triangle with two legs a, b and hypotenuse c When investigating a right-angled triangle (or shortly, right triangle), the following conclusions are often used: Theorem III triangle is a right triangle, if and only if the median on one side is half of the side Theorem IV If a right triangle has an interior angle of size 30 ±, then its opposite leg is half of the hypotenuse xamples xample 1 iven that the perimeter of a right angled triangle is (2 + p 6) cm, the median on the hypotenuse is 1 cm, find the area of the triangle Solution The Theorem III implies that = = = 1, so = 2 Let = b, = a, then 59

2 60 Lecture 10 Pythagoras Theorem and Its pplications a 2 + b 2 =2 2 =4 and a + b = p 6 Therefore 6 = (a + b) 2 = a 2 + b 2 +2ab, so ab = 6 4 = 1, 2 the area of the triangle is xample 2 s shown in the figure, \ = 90 ±, \1 =\2, =15 cm, =25 cm ind Solution rom introduce?, intersecting at When we fold up the plane that 4 lies along the line, then coincides with, so =, = =15 (cm) y applying Pythagoras Theorem to 4, = p 2 2 = p = 2(cm) 1 2 Letting = = x cm and applying Pythagoras Theorem to 4 leads the equation (x + 2) 2 = x , 4x = 12, ) x =3 Thus =3cm xample 3 s shown in the figure, is a square, P is an inner point such that P : P : P =1:2:3 ind \P in degrees Solution Without loss of generality, we assume that P =1,P =2,P = 3 Rotate the 4P around by 90 ± in clockwise direction, such that P! Q,!, then 4PQ is an isosceles right triangle, therefore PQ 2 =2P 2 =8,Q 2 = P 2 =1, therefore, by Pythagoras Theorem, P 2 =9=Q 2 + PQ 2, \QP = 90 ± Hence \P = \Q = 90 ± + 45 ± = 135 ± P Q

3 Lecture Notes on Mathematical Olympiad 61 xample 4 (SSSMO(J)/2003) The diagram shows a hexagon made up of five right-angled isosceles triangles O, O, O, O, O, and a triangle O, where O is the point of intersection of the lines and iven that O =8cm, find the area of 4O in cm 2 Solution rom O = p 1 2 O = ( p 1 2 ) 2 O = 1 2 O, O = p 1 2 O = 1 4O = 2(cm) Since Rt4O ª Rt4O, = O = 1 4 O = 1 4 p 2 O Let? at, then = 1 p 2 O = 1 8 O = 1 cm Thus, the area of 4O, S 4O, is given by O S 4O = 1 2 O = 4(cm2 ) xample 5 (ormula for median) In 4, M is the median on the side Prove that = 2(M 2 + M 2 ) Solution Suppose that? at y Pythagoras Theorem, 2 = =(M + M) = M 2 +2M M + M 2 + M 2 M 2 = M 2 + M 2 +2M M Similarly, we have 2 = M 2 + M 2 2M M M Thus, by adding the two equalities up, since M = M, = 2(M 2 + M 2 ) Note: When M is extended to such that is a parallelogram, then the formula of median is the same as the parallelogram rule: = xample 6 In the figure, \ = 90 ±, \ = 30 ±, is the mid-point of and?, =4cm ind

4 62 Lecture 10 Pythagoras Theorem and Its pplications Solution onnect Since is the perpendicular bisector of, =, so \ = \ = \ = 30 ±, \ = 60 ± 30 ± = 30 ±, ) = 1 2 = = 1 2 = 2 cm Now let = x cm, then from Pythagoras Theorem, (2x) 2 = x =) x 2 = 12 =) x = p 12 = 2 p 3 cm Thus, = 2 p 3 cm xample 7 or 4, O is an inner point, and,, are on,, respectively, such that O?, O?, and O? Prove that = Solution y applying the Pythagoras Theorem to the triangles O, O, O, O, O and O, it follows that = O 2 O 2 + O 2 O 2 + O 2 O 2 =(O 2 O 2 )+(O 2 O 2 )+(O 2 O 2 ) = The conclusion is proven O xample 8 In the diagram given below, P is an interior point of 4, PP 1?, P P 2?, PP 3?, and P 1 = P 2, P 2 = P 3, prove that P 1 = P 3 Solution or the quadrilateral P 1 P, since its two diagonals are perpendicular to each other, P1 2 + P 2 = 2 + P P 2 = P 2 + P1 2 y considering P 3 P and PP 2 respectively, it follows similarly that P 2 + P 2 3 = P P 2, P P 2 = P 2 + P 2 2 Then adding up the three equalities yields P 2 1 = P 2 3, ) P 1 = P 3 P 1 P P 2 P 3

5 Lecture Notes on Mathematical Olympiad 63 xample 9 In square, M is the midpoint of and N is the midpoint of M Prove that \N =2\M Solution Let = = = = a Let be the midpoint of Let the lines and intersect at y symmetry, we have = = a Since right triangles M and are symmetric in the line, \M = \ It suffices to show \N = \, and for this we only need to show \N = \N since \ = \ y assumption we have MN N = 3 4 a, ) N = r( 3 4 a)2 + a 2 = 5 4 a On the other hand, N = 1 4 a + a = 5 4 a, so N = N, hence \N = \N Testing Questions () 1 (HIN/1995) In 4, \ = 90 ±, =, is a point on Prove that =2 2 2 iven that Rt4 has a perimeter of 30 cm and an area of 30 cm 2 ind the lengths of its three sides 3 In the Rt4, \ = 90 ±, is the angle bisector of \ which intersects at iven = 15 cm, =9cm, : =5:3 ind the distance of from 4 In the right triangle, \ = 90 ±,= 12 cm, =6cm, the perpendicular bisector of intersects and at and respectively ind 5 In the rectangle,? at, = 1 4 and =5cm ind the length of

6 64 Lecture 10 Pythagoras Theorem and Its pplications 6 In 4, \ = 90 ±,is the mid-point of Prove that =4 2 7 In the right triangle, \ = 90 ±,, are points on and respectively Prove that = (HNMOL/1990) 4 is an isosceles triangle with = = 2 There are 100 points P 1,P 2,,P 100 on the side Write m i = P 2 i + P i P i (i =1, 2,,100), find the value of m 1 + m m In 4, \ = 90 ±, is the midpoint of,, are two points on and respectively, and? Prove that 2 = (HIN/1996) iven that P is an inner point of the equilateral triangle, such that P =2,P =2 p 3, P =4 ind the length of the side of 4 Testing Questions () 1 (SSSMO(J)/2003/Q8) is a chord in a circle with center O and radius 52 cm The point M divides the chord such that M = 63 cm and M = 33 cm ind the length OM in cm 2 (HIN/1996) is a rectangle, P is an inner point of the rectangle such that P =3,P =4,P =5, find P 3 etermine whether such a right-angled triangle exists: each side is an integer and one leg is a multiple of the other leg of the right angle 4 (HSM/1996) In rectangle, \ is trisected by and, where is on, is on, =6and =2 Which of the following is closest to the area of the rectangle? () 110, () 120, () 130, () 140, () (Hungary/1912) Let be a convex quadrilateral Prove that? if and only if = 2 + 2

7 Lecture 11 ongruence of Triangles Two triangles are called congruent if and only if their shapes and sizes are both the same In geometry, congruence of triangles is a very important and basic tool in proving the equality relations or inequality relations of two geometric elements (eg two segments, two angles, two sums of sides, two differences of angles, etc) Two triangles are congruent means they are the same in all aspects, so any corresponding geometric elements are equal also To prove two geometric elements being equal, it is convenient to construct two congruent triangles such that the two elements are the corresponding elements of the congruent triangles To prove two geometric elements are equal or not equal, even though their positions are wide apart, by using the congruence of two triangles, we can move the position of a triangle which carries one element, such that these two elements are positioned together so their comparison becomes much easier asic riteria for ongruence of Two Triangles (i) (ii) (iii) (iv) (v) SS: Two sides and their included angle of one triangle are equal to those in the other triangle correspondingly S: Two angles and one side of a triangle are equal to those in the other triangle correspondingly SSS: Three sides of a triangle are equal to those of the other triangle correspondingly or right triangles, these criteria can be simplified as follows: S: One side and one acute angle of a triangle are equal to those of the other triangle correspondingly SS: Two sides of a triangle are equal to those of the other triangle correspondingly 65

8 66 Lecture 11 ongruence of Triangles xamples xample 1 s shown in the diagram, given that in 4, =, is on and is on the extension of such that = The segment intersects at Prove that = Solution rom introduce k, intersecting at, as shown in the right diagram Then \ = \,\ = \ Since \ = \ = \, we have = = Therefore 4 ª = 4(S), hence = xample 2 iven and are the altitudes of the 4 P, Q are on and the extension of respectively such that P =, Q = Prove that P? Q Solution rom? Q and? \ = \Q Since = Q and P =, 4P ª = 4Q (SS), )\P = \Q, )\QP = \Q + \P = \Q + \Q = 180 ± 90 ± = 90 ± Q P O xample 3 (HIN/1992) In the equilateral 4, the points and are on and respectively, such that and intersect at P, and the area of the quadrilateral P is equal to area of 4P ind \P Solution rom and introduce? at and? at

9 Lecture Notes on Mathematical Olympiad 67 The condition [P ] =[P] implies that [] =[] Since =, so = Since \ = \, so Rt4 ª = Rt4 (S) therefore =, hence 4 ª = 4(SS) Thus, \ = \, so that \P = \P + \P = \P+ \P = 60 ± P xample 4 (HIN/1991) iven that is an equilateral triangle of side 1, 4 is isosceles with = and \ = 120 ± If points M and N are on and respectively such that \MN = 60 ±, find the perimeter of 4MN Solution *\ = \ = 30 ±, )?,? xtending to P such that P = N, then 4N ª = 4P (SS), therefore P = N \PM = 60 ± = \MN implies that 4PM ª = 4MN, (SS), ) PM = MN, ) MN = PM = M + PM = M + N Thus, the perimeter of 4MN is 2 P M N Note: Here the congruence 4PM ª = 4MN is obtained by rotating 4N to the position of 4P essentially xample 5 s shown in the figure, in 4, is the mid-point of, \ = 90 ±, intersects at and intersects at Prove that + > Solution In this problem, for the comparison of + and it is needed to move the segments,, together in a same triangle, and constructing congruent triangles can complete this task as follows Rotate 4 around by 180 ± in clockwise direction, then

10 68 Lecture 11 ongruence of Triangles!,! onnect,, Since = and =, we have 4 ª = 4, (SS), hence = < + = + xample 6 (HIN/1999) iven that 4 is a right-angled isosceles triangle with \ = 90 ± is the mid-point of, is perpendicular to, intersecting and at and respectively Prove that \ = \ Solution It is inconvenient to compare \ and \ directly To change the position of \, suppose that the perpendicular line from to intersects the line at Since =, \ = \ = 90 ± \, 4 ª = 4 (S), )\ = \ = \ * = = and \ = \ = 45 ±, ) 4 ª = 4 (SS), hence )\ = \ = \ xample 7 (HIN/1992) In the square, is the midpoint of, and intersect at Prove that? Solution Let be the point of intersection of and It suffices to show \ = \ y symmetry we have 4 ª = 4, 4 ª = 4, therefore \ = \ = \

11 Lecture Notes on Mathematical Olympiad 69 xample 8 (HIN/1992, 1993) In the graph, triangles and are both equilateral with a, b, c being collinear, M and N are midpoints of and respectively, intersects at and intersects at H Prove that (i) 4MN is equilateral, (ii) H k Solution (i) rom =, =,\ = \ = 120 ± 4 ª = 4 (SS), )\M = \N,M = N, which implies 4M ª = 4N (SS) ) M = N and \M = \N M \MN = \M + \M = \ = 60 ±, =) 4MN is equilateral N S T (ii) rom, H introduce S? at S and HT? at T respectively Since \ = \ = \H = \ = 60 ± =) k, H k, =) =,H = =) = H Since \S = \HT = 60 ±, so Rt4S ª = Rt4HT (S) Thus, S = HT, ie H k H Testing Questions () 1 In 4, \ = 60 ±, \ = 75 ±,? at,? at, intersects at H ind \H in degrees 2 4 is equilateral, is an inner point of 4 and P is a point outside 4 such that =, = P, and bisects \P ind \P 3 iven that the side of the square is 1, points P and Q are on and respectively, such that the perimeter of 4P Q is 2 ind \PQ in degrees by use of congruence of triangles 4 is a square, and are the midpoints of the sides and respectively If M is the point of intersection of and, prove that M =

12 70 Lecture 11 ongruence of Triangles 5 is a trapezium with k, \ = \ = 90 ±, and = = Prove that \ = 1 3 \ 6 (MOSOW/1952) In an isosceles triangle, =, \ = 20 ± M, N are on and respectively such that \M = 60 ±, \N = 50 ± ind \NM in degrees 7 iven that 4 is an isosceles right triangle with = and \ = 90 ± is a point on and is on the extension of such that? If = 1, prove that bisects \ 2 8 (HIN/1999) In the square, =8, Q is the midpoint of the side Let \Q = Æ On take a point P such that \P =2Æ If P = 10, find P 9 (HIN/1992) In the pentagon, \ = \ = 90 ±, = = = + =1 ind the area of 10 (NORTH UROP/2003) is an inner point of an equilateral 4 satisfying \ = 150 ± Prove that the triangle formed by taking the segments,, as its three sides is a right triangle Testing Questions () 1 (HIN/1996) iven that the segment is on a line ` On one side of ` take a point and construct two squares K and respectively outside the 4 Let M be the midpoint of the segment, prove that the position of M is independent of the choice of the position of 2 (HIN/1998) In Rt4, \ = 90 ±,? at, bisects \, intersects and at and respectively If is parallel to, intersecting at, prove that = 3 (HIN/1994) In 4, =2 and \ =2\ Prove that? 4 (HIN/2000) In a given quadrilateral, =, \ = 60 ±, \ = 120 ± Prove that + = 5 In 4, \ = \ = 80 ± The point P is on such that \P = 30 ± Prove that P =