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| | ==Solution== | | ==Solution== |
| − |
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| − | [asy]
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| − | size(10cm);
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| − | pair A=(0,0), B=(9,0), C=(15,8), D=(-6,8), C_1=(15,0), P=(9/2,12/5);
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| − | draw(B--C--D--A); draw(C--P--D); draw(B--C_1--C,dashed); draw((A+B)/2--(C+D)/2,dashed);
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| − | label("<math>A</math>",A,SW); label("<math>B</math>",B,S); label("<math>C'</math>",C_1,SE); label("<math>C</math>",C,NE); label("<math>D</math>",D,NW); label("<math>P</math>",P+(0.5,0),E);
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| − | draw(A--B--P--cycle,red+linewidth(1.1));
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| − | dot(A); dot(B); dot(C); dot(D); dot(P); dot(C_1);
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| − | label("<math>9</math>",(A+B)/2,dir(A--B)*dir(-90)); label("<math>10</math>",(B+C)/2,dir(B-C)*dir(90)); label("<math>6</math>",(B+C_1)/2,dir(B-C_1)*dir(90)); label("<math>8</math>",(C+C_1)/2,dir(C_1-C)*dir(90)); label("<math>21</math>",(C+D)/2,dir(C--D)*dir(-90)); label("<math>\frac{12}{5}</math>",(2*P+A+B)/4+(-0.1,-0.3),E);
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| − | [/asy]
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| − |
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| − | Note that <math>17^2-15^2=8^2=10^2-6^2</math> and <math>15-6=9</math> so if <math>C'</math> is the projection of <math>C</math> onto <math>AB</math> then <math>AC'=15,BC'=6,CC'=8</math>. It follows that <math>CD=21</math>. Let the diagonals <math>AC</math> and <math>BD</math> intersect at <math>P</math>, then <math>\triangle{PAB}\sim\triangle{PCD}</math> with similarity factor <math>\frac{9}{21}</math>. Thus the height of <math>\triangle{PAB}</math> is <math>\frac{9}{9+21}\cdot8=\frac{12}{5}</math> so <math>\left[PAB\right]=\frac{1}{2}\cdot9\cdot\frac{12}{5}=\frac{54}{5}</math> and hence the answer is <math>\boxed{059}</math> as desired.
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and 
, lie on the same side of line 
so that 
and 
are congruent with 
, 
, and 
. The intersection of these two triangular regions has area 
, where 
and 
are relatively prime positive integers. Find 
.
