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This week's Challenge Question:
Given a rectangle CFKE of any given sizes of 1 to 10, and located in a right angle triangle ABC, such that the hypotenuse AB touches the corner of the rectangle at E. Show how to construct a triangle ABC using only a compass and straight edge, no measuring instrument to be used. The hypotenuse length being any combination of sizes taken from the rectangle, the chosen length for the hypotenuse AB being greater than the length that would be a right angle with the rectangle diagonal CE, and referring to the construct, show how to calculate angle CAB, such that using the formula (FE/Sine A ) + (EK/ Cos A) = AB, show that AB is correct to within three decimal places.

Special thanks to jdretired who supplied this week's challenge.
And the Answer is....
We will present two solutions to this problem. There may be more solutions given the fact that this is a geometric construction problem.
SOLUTION 1:
Step 1. The solution concept outlines a number of rules. If a circle is drawn around the triangle ABC, from point O at the centre of AB, such that it touches all points, and a line CE extended to intersect the circle at D, and D used to form a triangle ABD, the triangle ABD is a simular triangle to CEF, and if a rectangle EGDJ is drawn within the triangle ABD, then the triangle EDG is a simular triangle to ABC. Also if line CD is taken as the base line, then the heights of points F and G are the same, both equal (h = 2.4). And lastly if lines AB, FK and GJ are extended, they all intersect at L, the position of L varying with length AB.
Note: The following solution holds true for any rectangle who's side measure any where between 1 and 10. And angle CEA is equal to or greater than 90 deg.

With the above concept rules in mind we construct the triangle ABD on a base line equal to the chosen length of A,B, in STEP ONE the chosen length A,B = 11, A,D = 8.8, B,D = 6.6, D,W = 5.28 and B,W = 3.96. We then solve by a number of feed back loops, to locate point C. this is done by establishing a number of POINTS, the location of a POINT is referenced by X, the horizontal distance from B, and Y, the vertical distance from B.
Step 2. Draw line B,N, being a tangent of the small circle of radius 2.4, then draw a second line parallel to B,N off set by 2.4 to intersect line A,B at P1. With compass set to Radius 3, from centre P1 intersect line B,N at Q1. With compass set to radius 4, from centre Q1 intersect the circle at point T1, draw a line from B to T1, and a line from P1 to Q1 which is extend it to intersect line B,T1 at POINT2.

MATHS
POINT1 X1 = 0. Y1 = 0
Angle D,B,N = Sine 2.4/6.6 = 21.323686 deg.
Angle N,B,P1 = 53.1301 - 21.32369 = 31.806416 deg. B,P1 = 2.4/Sine 31.80642 = 4.553639.
By Cosine rule ( 3, 4.553639, Cos 21.323686 ) .B,Q1 = 2.069836.
By Cosine rule ( 5.5, 2.069836, Cos 31.806416 ) .Q1,O = 3.896799.
(Q1,M)^2 = 4^2 - (5.5-x)^2 and (Q1,M)^2 = 3.896798^2 - x^2 therefore x = 2.67591.
Using x, angle Q1,O,M = 46.6309 deg.
By Cosine rule (5.5,3.896798, Cos O = 2.0669836) Cos O = 16.25724 deg.
Therefore angle O,B,T1 = (180 - (16.25724+46.6309) )/2 = 58.55593 deg.
By Sine rule (Point2,B,P1) B,POINT2 = 1.68204.
X2 = 1.68204 * Cos 58.55594 = 0.87746, and Y2 = 1.68204 * Sine 58.55594 = 1.43503
POINT2 X2 = 0.87746, Y2 = 1.43503
Step 3. Draw a line B1,N1, being a tangent at N of the small circle of radius 2.4 and passing through POINT2, draw a second line parallel to B1,N1 off set by 2.4 to intersect line AB at P2.

MATHS
POINT2,G = 3.96 - X2 = 3.08236. D,G = 5.28 - Y2 = 3.84497.
Distance between POINT2 and D = square root of ( 3.08236^2 + 3.84497^2) = 4.927955.
Angle D,POINT2,G = Cos 3.08236/4.927955 = 51.282232 deg
Angle D,POINT2,N = Sine 2.4/4.92795 = 29.144736 deg
Angle G,POINT2,N and angle POINT2,L,P2 = 51.282188 - 29.144766 = 22.137496 deg
Angle B,POINT2,L = 180-((180-58.55592)+22.137496) = 36.418424 deg
Distance between L,P2 = 2.4/Sine22.137496 = 6.36891.
By Sine rule (1.68204/Sine 22.137496 = L,B,/Sine 36.418424) L,B = 2.64997.
Therefore B,P2 = 6.36891 - 2.64997 = 3.718941.
Step 4. With compass set to Radius 3, from centre P2 intersect line B1,N1 at Q2. With compass set to radius 4, from centre Q2 intersect the circle at point T2, draw line B,T2, and line P2 to Q2 which is extend it to intersect line B,T2 at POINT3.

MATHS
Angle B,P2,Q2 = 90 - ( 22.137496 + 36.8699 ) = 30.992606 deg.
O,Q2 By Cosine rule(3,(5.5 - 3.718941),Cos(180-30.992606)) = 4.618752.
(Q2,M)^2 = 4^2 - (5.5-x)^2 and (Q2,M)^2 = 4.618752^2 - x^2 therefore x = 3.234807.
Using x, Q2,O,M = 45.543799 deg.
By sine rule [ Sine B,O,Q2/3 = Sine((180-30.992606)/4.618752) B,O,Q2 = 19.539633 deg.
Therefore angle B,O,T2 = 19.539633+45.543799 = 65.083432 deg.
Angle T2,B,P2 = (180 - 65.083432)/2 = 57.458284 deg.
Angle B,POINT3,P2 = 180 - ( 57.458284 + 30.992606) = 88.450891.
By Sine rule ( B,POINT3/Sine30.992606 = 3.718941/Sine 88.450891) B,POINT3 = 1.915685.
X2 = 1.915207 * Cos57.458284 = 1.030473, and Y2 = 1.915685 * Sine57.458284 = 1.61492.
POINT3 X3 = 1.030473, Y3 = 1.61492.
Step 5. With compass set to Radius 3, from centre POINT3 intersect line A,B at P3, with compass set to radius 5 (rectangle diagonal, CE ), from centre P3 intersect the circle at C. The lengths A,C and B,C will form a right angle triangle how's hypotenuse will touch a 3 by 4 rectangle at E.

MATHS
O,P3 =5.5-[ Square root(3^2 - Y3^2) + X3] = 1.94128.
By Cosine rule (5,5.5,Cos C ,(O,P3)) Cos C = 20.605454 deg
By Sine rule (SineO/5 = SineC/1.94128) SineO = 65.017921 deg
Therefore angle C,A,B = 65.017921/2 = 32.508961 deg.
CAB = 32.508961 deg, therefore by formula AB = 11.000.
SOLUTION 2:

With compass set at the selected hypotenuse length scribe an ark from point P1 to Q1, to dissect a vertical line extended from the rectangle, draw a line from Q1 to the corner of the rectangle at E, and extend it to dissect a point P2 on a line extended horizontally from the rectangle. With compass still at the selected hypotenuse length scribe an ark from point P2 to dissect the vertical line at Q2, draw a line from Q2 to the corner of the rectangle at E, and extend it to dissect the horizontal line at P3, repeat this a number of times that is practical, then transfer the final lengths C,Q and C,P and draw up triangle ABC.
MATHS
Base = Rectangle width
Start of Loop
Triangle height = square root ( Hypotenuse^2 - Base^2) .
Short height = Triangle height - rectangle height.
Top length = square root ( Short heigth^2 + rectangle width^2).
Bottom length = (rectangular height/short length) * Top length.
Length PQ = Top length + Bottom length.
Base = square root( Length PQ^2 - Triangle height^2).
With new Base size return to Start of loop four times,
Sine of angle CAB = Base/length PQ.
Hypotenuses check = (rectangle width/ Sine A) + ( rectangle height/ Cos A).
Examples
Width 4, height 3, length 11, CAB = 32.50575 deg. AB = 11.000.
Width 4.6, height 3.9, length 17, CAB = 21.024 deg. AB = 17.000
Width 8, height 3, length 33, CAB = 15.5263 deg. AB = 33.000
Width 5.8, height 9.2, length 30, CAB = 16.5144 deg. AB = 30.000
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Good Answers: