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Deep G
Member since Nov-6-05
Nov-06-05, 10:55 AM (EST)

"THe Isosceles Triangle problem"

 ABC is an isosceles triangle with angles A,B and C measuring 20,80 and 80 degrees respectively. Two rays BX and CY are drawn such that angle CBX measures 60 degrees and angle BCY measures 50 degrees. X and Y lie on sides AC and AB respectively.(i.e. the angles lie on the same side of side BC as the vertex A). Join XY. Find measure angle AXY.Curiously, this problem appeared to be very trivial...involving simplest of geometry (some angle measurements only)...but the solution eludes me.In fact, I thought of a solution to this problem ! Simply draw it and use the protractor !(hehe).. Any other way?

Subject     Author     Message Date     ID
THe Isosceles Triangle problem Deep G Nov-06-05 TOP
RE: THe Isosceles Triangle problem alexb Nov-08-05 1
RE: THe Isosceles Triangle problem mpdlc Nov-13-05 2
RE: THe Isosceles Triangle problem Deep G Nov-19-05 3
RE: THe Isosceles Triangle problem mr_homm Nov-28-05 4
RE: THe Isosceles Triangle problem mpdlc Nov-28-05 5
RE: THe Isosceles Triangle problem mr_homm Nov-28-05 6
RE: THe Isosceles Triangle problem mpdlc Nov-29-05 7
RE: The Isosceles Triangle problem Deep G Dec-03-05 8

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alexb
Charter Member
1697 posts
Nov-08-05, 10:56 AM (EST)

1. "RE: THe Isosceles Triangle problem"
In response to message #0

 Search this site for the "trigonometric form of Ceva's theorem".

mpdlc
guest
Nov-13-05, 08:12 AM (EST)

2. "RE: THe Isosceles Triangle problem"
In response to message #0

 I think this question is alike if not the same that the one below, that was treated in same extension there, by Stuart. I honestly did not try.The answer using trigonometry is a messy one, no strange in away to me it resembles to the two ladder crossing in a narrow alley problem, apparently easy at first sight but... ends in solving a quartic equation.Look first at the topic and if it not fits your request I will try to help.Conferences The CTK Exchange This and that Topic #636

Deep G
Member since Nov-6-05
Nov-19-05, 08:34 AM (EST)

3. "RE: THe Isosceles Triangle problem"
In response to message #2

 First of all, Thanx to both of you - alex and mpdlc for replying... :)but I purposely posted this particular problem (with these particular values) because it does (i think) have a geometric solution. The particular relationships that arise out of these angles do 'tease' one's cognitive prowess !ORIGINAL PROBLEM -ABC is an isosceles triangle with angles A,B and C measuring 20,80 and 80 degrees respectively. Two rays BX and CY are drawn such that angle CBX measures 60 degrees and angle BCY measures 50 degrees. X and Y lie on sides AC and AB respectively.(i.e. the angles lie on the same side of side BC as the vertex A). Join XY. Find measure angle AXY.I had it from a friend appearing for the Mathematics Olympiads(whose whereabouts i am not aware of right now :( ) six years ago. Here is wat i recall - (As you read ahead, take a look in the attached file)CONSTRUCTION - Draw angle CBZ 20 degrees with Z lying on AC. Join YZ. Let us jot down what do we have now -1. Triangle CBZ is isosceles. So, BC = BZ2. Triangle BCY is isosceles. So, BC = BY.1 and 2 give BY = BZ 3. Triangle BYZ is equilateral as BY = BZ and angle YBZ = 60 degrees. So, angle BYZ is 60 degrees. So we have : So angle CYZ is 10 degrees AND BC=BY=BZ=YZ.getting angle CYZ as 10 degrees was an intermediate step. Hereafter, he mentioned using one more construction of a circular arc.Can you help me ?
mr_homm
Member since May-22-05
Nov-28-05, 01:58 AM (EST)

4. "RE: THe Isosceles Triangle problem"
In response to message #3

 >First of all, Thanx to both of you - alex and mpdlc for >replying... :) >>but I purposely posted this particular problem (with these >particular values) because it does (i think) have a >geometric solution. The particular relationships that arise >out of these angles do 'tease' one's cognitive prowess ! Hi DeepG,Using the same method I used in the thread here,I got a value of the angle as the solution tosin(110-x)/sin(x) = sin(30)sin(60)/(sin(20)sin(40)), which gives x = 25.10437 degrees. Just as with the triangle problem in the other thread, this means that this angle is not one you can construct easily from the given angles (20,80,50,60,180). Therefore, I am forced to conclude that someone has misled you. This problem is NOT geometrically solvable with angles of 50 and 60 degrees, for exactly the same reason that the other problem is not solvable for 20 and 30 degrees. The trigonometric solution is not too hard (see above), but it is not the kind of solution you had hoped for.Sorry for the bad news,--Stuart Anderson

mpdlc
guest
Nov-28-05, 08:49 AM (EST)

5. "RE: THe Isosceles Triangle problem"
In response to message #0

 After being at sea for a week I workout the problem using analytic geometry to get the slope of line between point X Y I called P and Q just not to be confused with the customary name for the coordinate axis.The obvious procedure for me is 1) Locate the origin at the midpoint of the small side of the triangle the height of the triangle will be contained in OY axis. Obviously coordinates for B will be (w, 0) and for C (0, -w)2) Write the four equations for the two sides and the two straight lines BP and CQ, calling the slope Mb, Mc Bp and Cq.3) Solving the equations of one side AC with the straight line BP and AB with CQ will render the coordinates for P and Q. Now subtracting ordinates and abscises P from Q and dividing you get the slope of line PQ, plugging the values of Mb, Mc Bp and Cq we will get the angle now you we add the 80 degrees and have the solution.I got 110 degrees but since there is nothing interesting in the procedure other than algebra manipulation, nor I posted nor I think is worthy. Unless that you were trying to solve the problem for a very general case lets say making variable any of this above slope to determine the envelope of PQ, or imposing a kind of relation between them like have the point A moving in a circumference so the segment BC will be seen with a constant angle or other useless intricacies we were forced to solve in my old days at School of Naval Architecture.So I focus in one more pure geometry solution like you suggest. And below is the procedure I follow to get our elusive angle.Since we are not going to use analytic geometry, we can keep your original notation1) To keep the drawing tidy erase the line CY forming the 50 degree angle with the base.Them as an auxiliary line we will draw the height of our triangle emanating from A, we will call AH2) Construct the symmetrical lines to BX with respect to AH. We will call CX2. Also we will name H1 the point of intersection between BX and CX23) We will also draw line XX2. Obviously the triangle XX2H1 is equilateral. 4) Emanating from X an X2 we will draw two height of the mentioned triangle extending them till sides AB and AC, we will call R1 and R2 the points of intersection. 5) Draw a circumference passing for CBRR2 since they are concyclical. 6) Draw from R a parallel line to BX and do the same for R2 . These parallel lines will meet at AH in a point we will call O which is the center of circumference containing CBRR2. (Triangle RR2O is equilateral too) 7) From the point C of the circumference CBRR2 we will see the segment RR2 under 30 degree angle which means that angle R2CB is 50 degree so R2 is the same point of Y. Obviously the angle we try to find will measure (80+30) Sorry for using R2 and X2 but when I saw the preview my ' look got transformed in a question mark

mr_homm
Member since May-22-05
Nov-28-05, 05:01 PM (EST)

6. "RE: THe Isosceles Triangle problem"
In response to message #5

 Hi mpdlc,The angle I was trying to find is BXY, while you found AXY. If your answer is correct, then BXY should be 30 degrees, while I got approximately (but not exactly) 25 degrees. One of us is wrong!I also have a question about your step 6. I drew the diagram according to your instructions, and I understand all the steps, and the picture I got makes sense. I also agree with all the other steps, but in step 6 I do not see how you know that the meeting point of the lines through R and R2 is at the center of the circle CBRR2. It is clear that RR2O is equilateral, and that O is on the vertical axis of symmetry of the circle, but I just do not see why it is at the center.Thanks,--Stuart Anderson

mpdlc
guest
Nov-29-05, 11:50 AM (EST)

7. "RE: THe Isosceles Triangle problem"
In response to message #6

 Hi Stuart,I am going to give you the reasoning behind I assumed O is the center of the circumference CBR2R, it is more a sort a quick check than a formal proof.1) Angle RR2B equal 100 degrees since opposed angle in a isosceles trapezoid must add to 180 and angle in C equals 80.2) As RR2O is 60 then OR2B is 40, since I assume OR2 and OB are the radius angle R2BO is also 40 so the central angle BOR2 is 100.3) Same numbers for symmetrical isosceles triangle COR, (central angle COR equals 100). The above leads us to the triangle COB is also identical to the others OR2B and COR central angle (COB 100) . Angle ROR2 must be 60 as supposed by hipothesis and the important thing O is equidistant to R,R2, C and B. The equality of the three isosceles triangle get also validated taking under consideration the value of 50 degree for the angle BCY in the original set.Of course this is not a rigorous proof. I believe the right way to do it'should be assuming a different point call it M as a center for the circumference CBR2R and then working for �reductio ab absurdum � so M must be O. But Stuart� I am just an end user of math, .. do not ask me for more now I feel I got the flu.

Deep G
Member since Nov-6-05
Dec-03-05, 03:17 AM (EST)

8. "RE: The Isosceles Triangle problem"
In response to message #0

 Hi all,Its really frustrating that I have not been able to go through you people's replies because of my busy schedule. I have newly joined a software firm as a computer science engineer and work round the clock. But I look forward to a healthy and fruitful association with you all through CTK, thanks to Mr. Alexander Bogomolny ! Waiting impatiently to join "the league of extraordinary gentlemen"...Yours sincerely,Deepak Gupta.(Deep G)d:-)