Results 1 to 5 of 5

Math Help - Triangle proof

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Triangle proof

    ABC is a triangle with no obtuse angles. The perpendicular AD from A to BC meets the circumcircle of the triangle at E. If the incentre of the triangle is equidistant from A and E, prove that \cos B+\cos C=1.
    I have no idea how to begin.
    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Quote Originally Posted by arze View Post
    ABC is a triangle with no obtuse angles. The perpendicular AD from A to BC meets the circumcircle of the triangle at E. If the incentre of the triangle is equidistant from A and E, prove that \cos B+\cos C=1.
    I have no idea how to begin.
    thanks
    If I is the incenter, and AI = IE, then I must lie on the perpendicular bisector of AE and it must pass through circumcenter of ABC.
    Secondly I must be equidistant from all the three sides of the triangle ABC.
    These two conditions are satisfied only when ABC is equilateral triangle. In that case B = C = 60 degrees and cosB + cosC = 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello arze
    Quote Originally Posted by arze View Post
    ABC is a triangle with no obtuse angles. The perpendicular AD from A to BC meets the circumcircle of the triangle at E. If the incentre of the triangle is equidistant from A and E, prove that \cos B+\cos C=1.
    I have no idea how to begin.
    thanks
    This is much, much harder than the other trig questions you've posted. I wonder where it came from.

    I'm afraid this
    Quote Originally Posted by sa-ri-ga-ma View Post
    If I is the incenter, and AI = IE, then I must lie on the perpendicular bisector of AE and it must pass through circumcenter of ABC.
    Secondly I must be equidistant from all the three sides of the triangle ABC.
    These two conditions are satisfied only when ABC is equilateral triangle. In that case B = C = 60 degrees and cosB + cosC = 1.
    isn't true. The triangle ABC needn't be equilateral.

    I have a proof - see below - but it's very complicated for such an apparently simple result. Perhaps someone may be able to come up with something more straightforward?

    Here goes.

    Look at the attached diagram. I is the incentre.

    I haven't drawn the circumcircle - the only property of this that I'm using is that:
    \angle AEC = \angle ABC (angles in same segment) Call this equation (1)
    Then we need to express AE in two ways:
    1. In terms of the sides and angles of any triangle ABC.

    2. Using the fact that \triangle AEI is isosceles, with AI = IE.
    So first:
    AD= b \sin C ...(2)

    DC = b \cos C
    ...(3)

    DE = DC \cot \angle DEC = DC \cot B, from
    (1)
    = \frac{b\cos C \cos B}{\sin B}, from (3)
    \Rightarrow AE = AD +DE
    = b\sin C + \frac{b\cos C \cos B}{\sin B}, from (2)

     = \frac{b\cos(B-C)}{\sin B}, using \cos (B-C) = \cos B \cos C +\sin B \sin C. Call this equation (4)
    Then, expressing AE using the fact that \triangle AEI is isosceles:

    The radius of the incircle, r, is given by:
    r = AK =\frac{2\Delta}{a+b+c}, where \Delta is the area of the triangle ABC
    Now AI bisects \angle BAC. So \angle IAK = \tfrac12A
    \Rightarrow AI = \frac{AK}{\sin\tfrac12A}
    =\frac{2\Delta}{(a+b+c)\sin\tfrac12A} ...(5)
    Also, from \triangle ABD,\; \angle BAD = 90^o-B
    \Rightarrow \angle JAI = \tfrac12A - (90^o-B)
    =\tfrac12A+B-90^o
    So AE = 2AI\cos\angle JAI
    =2AI \cos(\tfrac12A+B-90^o)

    =2AI\sin(\tfrac12A+B)

    = \frac{4\Delta\sin(\tfrac12A+B)}{(a+b+c)\sin\tfrac1  2A}, from
    (5)
    So, from (4), equating these two expressions for AE, using (4):
    \frac{b\cos(B-C)}{\sin B}=\frac{4\Delta\sin(\tfrac12A+B)}{(a+b+c)\sin\tfr  ac12A}
    We now use \Delta = \tfrac12bc\sin A, and then use the Sine Rule to write a and b in terms of c:
    \Rightarrow\frac{b\cos(B-C)}{\sin B}=\frac{2bc\sin A\sin(\tfrac12A+B)}{(\frac{c\sin A}{\sin C}+\frac{c\sin B}{\sin C}+c)\sin\tfrac12A}
    =\frac{2b\sin A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)\sin\tfrac12A}
    \Rightarrow\frac{\cos(B-C)}{\sin B}=\frac{4\sin\tfrac12 A\cos\tfrac12A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)\sin\tfrac12A}
    =\frac{4\cos\tfrac12A\sin C\sin(\tfrac12A+B)}{(\sin A+\sin B+\sin C)}
    All that remains now - and there's still quite a bit to do - is to eliminate A using A = 180^o-(B+C), and the result should follow. So, noting that:
    \cos\tfrac12A = \cos(90-\tfrac12[B+C])
    =\sin\tfrac12(B+C)
    and
    \sin(\tfrac12A+B) = \sin(90-\tfrac[B+C]+B)
    =\cos\tfrac12(B-C)
    and
    \sin A = \sin (B+C)
    we get:
    \frac{\cos(B-C)}{\sin B}=\frac{4\sin\tfrac12(B+C)\sin C\cos\tfrac12(B-C)}{\sin(B+C)+\sin B + \sin C}
    =\frac{4\sin\tfrac12(B+C)\sin C\cos\tfrac12(B-C)}{2\sin\tfrac12(B+C)\cos\tfrac12(B+C)+2\sin\tfra  c12(B + C)\cos\tfrac12(B-C)}

    =\frac{2\sin C\cos\tfrac12(B-C)}{\cos\tfrac12(B+C)+\cos\tfrac12(B-C)}

    =\frac{4\sin\tfrac12 C\cos\tfrac12C\cos\tfrac12(B-C)}{2\cos\tfrac12B\cos\tfrac12C}
    \Rightarrow \frac{2\cos^2\tfrac12(B-C)-1}{2\sin\tfrac12B\cos\tfrac12B}=\frac{2\sin\tfrac1  2C\cos\tfrac12(B-C)}{\cos\tfrac12B}

     \Rightarrow 2\cos^2\tfrac12(B-C)-1=4\sin\tfrac12B\sin\tfrac12C\cos\tfrac12(B-C)
     \Rightarrow 2\cos\tfrac12(B-C)\Big(\cos\tfrac12(B-C)-2\sin\tfrac12B\sin\tfrac12C\Big)=1

    \Rightarrow2\cos\tfrac12(B-C)(\cos\tfrac12B\cos\tfrac12C+\sin\tfrac12B\sin\tf  rac12C-2\sin\tfrac12B\sin\tfrac12C)=1

    \Rightarrow2\cos\tfrac12(B-C)(\cos\tfrac12B\cos\tfrac12C-\sin\tfrac12B\sin\tfrac12C)=1

    \Rightarrow2\cos\tfrac12(B-C)\cos\tfrac12(B+C)=1

    \Rightarrow \cos B + \cos C = 1
    Phew!

    You would think that there was an easier way, but I can't see it!

    Grandad
    Attached Thumbnails Attached Thumbnails Triangle proof-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    I also found this a tough problem. I have a cumbersome solution that I held back from posting in the hope that someone would find a neater way of doing it. No offence to Grandad, but his solution is about as cumbersome as mine. Just for the record, here's an outline of how I did it.

    The incentre I is on the perpendicular bisector of AE. So is the circumcentre O. So the line IO is parallel to BC (see the attached modified version of Grandad's diagram). My proof is based on the fact that the triangles BOC and BIC have the same base BC and the same vertical height, and therefore the same area.

    The area of BOC is \tfrac12aR\cos A, where R is the circumradius and a is the length BC. But \frac a{\sin A} = 2R. Also, \sin A = \sin(B+C) and \cos A = -\cos(B+C). So the area of triangle BOC is -\tfrac14a^2\frac{\cos(B+C)}{\sin(B+C)}.

    The area of BIC is \tfrac12IB.IC\sin(\angle BIC). By the sine rule, \frac{BI}{\sin\frac12C} = \frac{CI}{\sin\frac12B} = \frac{a}{\sin\frac12(B+C)}. So the area of triangle BIC is \tfrac12a^2\frac{\sin\frac12B\sin\frac12C}{\sin\fr  ac12(B+C)}.

    Equate those two areas to see that \frac{-\cos(B+C)}{\sin(B+C)} = \frac{2\sin\frac12B\sin\frac12C}{\sin\frac12(B+C)}. On the left side, the numerator is \sin B\sin C - \cos B\cos C and the denominator is 2\sin\tfrac12(B+C)\cos\tfrac12(B+C) = 2\sin\tfrac12(B+C)\bigl(\cos\tfrac12 B\cos\tfrac12 C - \sin\tfrac12 B\sin \tfrac12C\bigr). Make those substitutions and cross-multiply (noticing that \sin\tfrac12(B+C) cancels in the two denominators):

    \begin{aligned}\sin B\sin C - \cos B\cos C &= 4\sin\tfrac12B\sin\tfrac12C\bigl(\cos\tfrac12 B\cos\tfrac12 C - \sin\tfrac12 B\sin \tfrac12C\bigr)\\  &= \sin B\sin C -4\sin^2\tfrac12 B\sin^2 \tfrac12C.\end{aligned}

    Now use the fact that 2\sin^2\theta = 1-\cos2\theta to write this as \cos B\cos C = (1-\cos B)(1-\cos C), which simplifies to \cos B + \cos C = 1.

    Phew!
    Attached Thumbnails Attached Thumbnails Triangle proof-triangle.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    Thanks everyone! The proofs were really long.
    Grandad, I got this from an A-level pure math textbook.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Right Triangle Proof
    Posted in the Geometry Forum
    Replies: 4
    Last Post: November 2nd 2010, 06:35 PM
  2. Triangle - proof
    Posted in the Geometry Forum
    Replies: 1
    Last Post: November 1st 2010, 06:27 AM
  3. Proof in a triangle.
    Posted in the Geometry Forum
    Replies: 4
    Last Post: February 6th 2010, 06:29 AM
  4. triangle proof
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 1st 2008, 03:41 PM
  5. Plz Help me with this triangle proof
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: November 25th 2007, 04:12 AM

Search Tags


/mathhelpforum @mathhelpforum