Answer to Question #51274 in Trigonometry for Mr. X

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Answer to Question #51274 in Trigonometry for Mr. X

Question #51274

Hexagon ABCDEF is inscribed in the circle with centre M and radius R. AB=CD=EF=R. Points I, J and K are the midpoints of segments BC, DE and FA respectively. Prove that triangle IJK is equilateral.

Note: Could you please use trigonometry? The method is to find MI and MJ, and then use cosine rule in triangle MIJ to find IJ. Then simplify the expression obtained for IJ, and as it will be symmetrical, the triangle will be equilateral.

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Mr. X

16.03.15, 19:36

Oh, okay

Assignment Expert

16.03.15, 19:33

The statement was proved for a regular hexagon, more general case is unknown.

Mr. X

13.03.15, 19:42

Yes, but the hexagon is regular if and only if all the sides are equal to the radius. In this case, only alternate sides are given to be equal. Can it be proved that triangle IJK is equilateral by the given method?

Assignment Expert

13.03.15, 18:24

Dear Mr. X. Thank you for adding information.

Assignment Expert

13.03.15, 18:22

Dear Mr. X. You are right. Note that the regular hexagon has a property that the length of radius is equal to the length of side.

Mr. X

13.03.15, 17:08

This is the method the question says to follow. Let angle IMC=u, JMD=v, KMF=w. In triangle MIC, we get MI=Rcos(u) and in triangle MJD, we get MJ=Rcos(v). Now, in triangle MIJ, angle CMD will be 60 degrees, because it is an equilateral triangle, as you have shown above. So by using cosine rule, we get IJ^2=R^2 cos^2(u) + R^2 cos^2(v) + -2(Rcos(u))(Rcos(v))(cos(60+u+v)) Now I need to simplify this expression, and make it symmetrical in u, v, w, and R so that we can say that the triangle is equilateral.

Mr. X

13.03.15, 17:02

Yes, no vectors are used. What I meant is this. You have directly stated the following statement, in the 11th line, that angle BMC=180 degrees-AMB-CMD=(180-60-60)=60 degrees. By this statement, you are assuming that angles AMB, BMC and CMD sum to 180 degrees, which means that you are assuming that AMD is a straight line. But no such information is given in the question, as A, M and D are not necessarily collinear.

Assignment Expert

13.03.15, 16:43

We did not use collinear vectors in this proof, points A,M, D do not necessarily lie in the straight line. It is essential to prove the congruence of triangles, show that triangles are equilateral in order to deduce that hexagon ABCDEF is regular.

Assignment Expert

13.03.15, 15:49

Dear Mr. X, You're welcome. We are glad to be helpful. If you liked our service please press like-button beside answer field. Thank you!

Mr. X

12.03.15, 19:21

But in the proof that ABCDEF is a regular hexagon, you are assuming that A, M and D are collinear. No such information is provided by the question.

Mr. X

12.03.15, 18:50

Oh right, thanks! :)

Assignment Expert

12.03.15, 17:57

The statement about regular hexagon can be deduced. It follows that AM=BM=CM=DM=R from the fact that hexagon ABCDEF is inscribed in the circle with center M and radius R. If AB=CD=R and AM=BM=CM=DM=R, then AB=AM=BM=R, CD=CM=DM=R, hence 1) triangles ABM and CDM are congruent because they are equal in three sides; 2) triangles ABM and CDM are equilateral, because they have three sides of equal length. Reasonings 1) give that a) BM=CM, hence triangle BMC is iscosceles and angle CBM=BCM. Reasonings 2) give that b) angle AMB=60 degrees, angle CMD=60 degrees, hence taking into account the fact that the sum of the angles in triangle is 180, obtain angle BMC=180 degrees-AMB-CMD=(180-60-60)=60 degrees. Taking into account CBM=BCM from a), BMC=60 degrees from b), the fact that the sum of the angles in triangle is 180, come to the following system of equations CBM=BCM, BMC=60, CBM+BCM+BMC=180, which is reduced to 2CBM=180-60, hence CBM=60 and finally CBM=BCM=BMC=60. The last result means that triangle BMC is equilateral, hence BM=CM=BC=R. Thus, given AB=CD=R, it was shown that AB=BC=CD=R in this problem. Similarly you can prove other equalities: AB=BC=CD=DE=EF=AF=R. It means that hexagon ABCDEF is regular.

Mr. X

11.03.15, 19:21

It is not given that the hexagon is regular. It is only given that AB=CD=EF=R.

Answer to Question #51274 in Trigonometry for Mr. X

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