Emergency problems #4

Emergency problems #4 - MATH 348: Solved Emergency...

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MATH 348: Solved Emergency Problems, Week 4 Leonid Chindelevitch August 26, 2008 1 Monday, July 28, 2008 1. Problem: From each of the centers O 1 ,O 2 of two circles ω 1 2 tangents are drawn to the other circle. Prove that equal chords are intercepted on the circumferences. Solution: Let O 1 E,O 1 F be the tangents to ω 2 , let O 2 G,O 2 H be the tangents to ω 1 , and let A,B and C,D be the intersections of these tangents with ω 1 and ω 2 , respectively. Let M and N be the midpoints of AB and CD , respectively. The line O 1 O 2 is the bisector of the angles AO 1 B and CO 2 D , so that M and N lie on it. The right-angled triangles O 1 EO 2 and O 1 MA are similar by two angles, and so are O 1 GO 2 and CNO 2 . We deduce that EO 2 O 1 O 2 = MA O 1 A ; O 1 G O 1 O 2 = CN CO 2 . From these equalities, we now get AB = 2 AM = 2 O 1 A · O 2 E O 1 O 2 = 2 O 1 G · O 2 C O 1 O 2 = 2 CN = CD. It follows that the two chords, AB and CD , are indeed equal. 1
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2. Problem: Prove that if perpendiculars are drawn from the feet of the altitudes of a triangle to each of the other two sides, the six feet of these perpendiculars lie on a circle. Solution: This circle is known as the Taylor circle of a triangle. For a solution, see the National Mathematics Magazine, Vol. 18, No. 1 (Oct. 1943), p.40, or email me. 3. Problem: Prove that if, in a triangle, two medians are perpendicular, then the three medians are the sides of a right-angled triangle. Solution: Let Δ ABC be the given triangle, let D,E,F be the midpoints of the sides a,b,c respectively, and let G be the centroid. Let us define x := GD,y := GE ; since the centroid divides the medians in a 1:2 ratio, we have GA = 2 x,GB = 2 y . Assume that the medians AD and BE are perpendicular; by Pythagoras’ theorem in the triangles AGB,AGE,BGD , we obtain: (2 x ) 2 + (2 y ) 2 = c 2 ; x 2 + (2 y ) 2 = ( a/ 2) 2 ; (2 x ) 2 + y 2 = ( b/ 2) 2 . From the formula for the length of the median proven in practice problem
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Emergency problems #4 - MATH 348: Solved Emergency...

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