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Let $R_1$ be the circumradius of triangle $ABD,$ and let $R_2$ be the circumradius of triangle $BCD.$ Compute $\frac{R_1}{R_2}.$

[asy] unitsize(1 cm);pair A, B, C, D;A = (-2,0);B = (0,5);C = (2,0);D = (-1,0);draw(A--B--C--cycle);draw(B--D);label(

 

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 Apr 13, 2022
 #1
avatar+122390 
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I don't know the most direct method of this.....but

 

The circumradius  of a triangle is given by    :     (product of the sides) / [  4 * area ]

 

Since the two interior triangles are under the same  height, they are each other as to their bases

 

So area of   BDC =  3 area of  BDA

 

So, since BD is common to both,  the product of the sides of  BDC  = 3 times the product of the sides of  BDA

 

So

 

R1  / R2  =      [  product of the sides of BDA ] / [ 4 *area of BDA ]

                      ________________________________________________     =

                       [ 3 * product of the sides of  BDA ] / [ 4 * (3 *area of BDA ) ]

 

 

 

[ product of the sides of BDA ]     *  [ 4*( 3 * area of BDA) ] 

______________________________________________  =    12 / 12  =   1

[ 3 * product of the sides of BDA]  * [ 4 *area of BDA ]

 

 

cool cool cool

 Apr 13, 2022

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