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Math Help - rectangle

  1. #1
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    rectangle

    Let ABCD be a rectangle with BC=3AB .
    Let P and Q are two points on side BC with PB=PQ=QC .

    Show that

    < DPC +<DBC =< DQC .
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  2. #2
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    Hello,

    Quote Originally Posted by perash View Post
    Let ABCD be a rectangle with BC=3AB .
    Let P and Q are two points on side BC with PB=PQ=QC .

    Show that

    < DPC +<DBC =< DQC .
    For that, I will use a rough method, using more algebra than geometry.

    ----------------------------
    Because I don't remember the formula for \arctan a+\arctan b, I'll show it.

    ~~~~~~~~~~~~~~
    For that I will need to find the formula (which I don't remember either) of tan(c+d) :
    \tan(c+d)=\frac{\sin(c+d)}{\cos(c+d)}

    \tan(c+d)=\frac{\sin(c)\cos(d)+\cos(c)\sin(d)}{\co  s(c)\cos(d)-\sin(c)\sin(d)}

    Let's multiply by \frac{~~~~\frac{1}{cos(c)\cos(d)}~~~~}{~~~~\frac{1  }{cos(c)\cos(d)}~~~~}

    This gives :

    \tan(c+d)=\frac{\tfrac{\sin(c)}{\cos(c)}+\tfrac{\s  in(d)}{\cos(d)}}{1-\tfrac{\sin(c)}{\cos(c)} \cdot \tfrac{\sin(d)}{\cos(d)}}

    \boxed{\tan(c+d)=\frac{\tan(c)+\tan(d)}{1-\tan(c)\tan(d)}}

    ~~~~~~~~~~~~~~

    Therefore, we have :

    \tan(\arctan(a)+\arctan(b))=\frac{\tan(\arctan(a))  +\tan(\arctan(b))}{1-\tan(\arctan(a))\tan(\arctan(b))}

    \tan(\arctan(a)+\arctan(b))=\frac{a+b}{1-ab}


    \implies \boxed{\arctan(a)+\arctan(b)=\arctan\left(\frac{a+  b}{1-ab}\right)}

    ----------------------------

    Now let's go back to the problem.

    rectangle-rectangle.jpg

    So we have \angle DPC+\angle DBC=\arctan(1/3)+\arctan(1/2)

    \arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{1/3+1/2}{1-(1/2)*(1/3)}\right) (formula above).

    \arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{5/6}{1-1/6}\right)=\boxed{\arctan(1)}



    Therefore, \boxed{\angle DPC+\angle DBC=\angle DQC}

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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,


    For that, I will use a rough method, using more algebra than geometry.

    ----------------------------
    Because I don't remember the formula for \arctan a+\arctan b, I'll show it.

    ~~~~~~~~~~~~~~
    For that I will need to find the formula (which I don't remember either) of tan(c+d) :
    \tan(c+d)=\frac{\sin(c+d)}{\cos(c+d)}

    \tan(c+d)=\frac{\sin(c)\cos(d)+\cos(c)\sin(d)}{\co  s(c)\cos(d)-\sin(c)\sin(d)}

    Let's multiply by \frac{~~~~\frac{1}{cos(c)\cos(d)}~~~~}{~~~~\frac{1  }{cos(c)\cos(d)}~~~~}

    This gives :

    \tan(c+d)=\frac{\tfrac{\sin(c)}{\cos(c)}+\tfrac{\s  in(d)}{\cos(d)}}{1-\tfrac{\sin(c)}{\cos(c)} \cdot \tfrac{\sin(d)}{\cos(d)}}

    \boxed{\tan(c+d)=\frac{\tan(c)+\tan(d)}{1-\tan(c)\tan(d)}}

    ~~~~~~~~~~~~~~

    Therefore, we have :

    \tan(\arctan(a)+\arctan(b))=\frac{\tan(\arctan(a))  +\tan(\arctan(b))}{1-\tan(\arctan(a))\tan(\arctan(b))}

    \tan(\arctan(a)+\arctan(b))=\frac{a+b}{1-ab}


    \implies \boxed{\arctan(a)+\arctan(b)=\arctan\left(\frac{a+  b}{1-ab}\right)}

    ----------------------------

    Now let's go back to the problem.

    Click image for larger version. 

Name:	rectangle.JPG 
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    So we have \angle DPC+\angle DBC=\arctan(1/3)+\arctan(1/2)

    \arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{1/3+1/2}{1-(1/2)*(1/3)}\right) (formula above).

    \arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{5/6}{1-1/6}\right)=\boxed{\arctan(1)}



    Therefore, \boxed{\angle DPC+\angle DBC=\angle DQC}

    . Why go to the arctans? The tans will do.

    If (angle DPC) +(angle DBC) = angle DQC
    Then, tan[<DPC +<DBC) = tan(<DQC)

    [tan(<DPC) +tan(<DBC)] /[1 -tan(<DPC)*tan(<DBC)] =? tan(<DQC)
    [1/3 +1/2] /[1 -(1/3)(1/2)] =? 1
    [5/6] /[5/6] =? 1
    1 =? 1
    Yes, so <DPC +<DBC = <DQC
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  4. #4
    Moo
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    Quote Originally Posted by ticbol View Post
    . Why go to the arctans? The tans will do.
    Because I imagined it this way on my sketch

    Thank you for poiting that out
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  5. #5
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    Proof using geometry

    As < DQC = 45, what we need to prove is that:
    < DPC + < DBC = 45
    Construct another rectangle BCEF with CE = BF = AB.
    R and S are points on EF such that FR = RS = SE.

    Connect B and S.
    triangle BQS triangle DCP are congruent, so
    < SBQ = < DPC
    Therefore
    < DPC + < DBC = < DBS

    Next we will show that < DBS = 45.

    In fact, triangle BSD is an isosceles right triangle with BS = DS and < BSD = 90. Its proof is as follows:

    As triangle BFS and triangle DES are congruent,
    BS = DS, and
    < BSF = < SDE

    Because QS and DE are parallel,
    < QSD = < SDE

    Thus
    < BSD = < BSQ + < QSD
    = < BSQ + < BSF
    = < FSQ
    = 90
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