document.write( "Question 1107405: My friend made up this geometry problem:
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Algebra.Com's Answer #722460 by ikleyn(53765) $\"\"$ $\"About$

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document.write( "In the Figure, you have three circles: Large, middle size and small circle A.\r\n" );
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document.write( "        Let \"a\" be the radius of the small circle A, which is under the question.\r\n" );
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document.write( "    Draw vertical radius of the Large circle from its center.\r\n" );
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document.write( "    Connect the centers of the small and the middle size circles.\r\n" );
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document.write( "    Draw the horizontal segment from the center of the small circle till the intersection with the vertical radius of the Large circle.\r\n" );
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document.write( "    Let H be the length of this horizontal segment.\r\n" );
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document.write( "    You get right-angled triangle with the legs H, 2-a and the hypotenuse 2+a.\r\n" );
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document.write( "    Hence, you have a Pythagorean equation\r\n" );
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document.write( "         =  + .     (1)\r\n" );
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document.write( "    Next, draw the radius of the Large circle from its center through the center of the small circle A.\r\n" );
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document.write( "    Draw the vertical segment from the center of the small circle A to the horizontal diameter of the Large circle.\r\n" );
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document.write( "    You will get right-angled triangle with the legs \"a\", H and the hypotenuse (2-a).\r\n" );
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document.write( "    Hence, you have second Pythagorean equation\r\n" );
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document.write( "         =  + .         (2)\r\n" );
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document.write( "    OK. The setup is completed, and you have now two equations for two unknowns\r\n" );
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document.write( "         =  + .     (1)\r\n" );
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document.write( "         =  + .         (2)\r\n" );
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document.write( "    Subtract eq(2) from eq(1) to eliminate H.  You will get a single equation for \"a\"\r\n" );
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document.write( "         = ,\r\n" );
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document.write( "        2a*2 =   ====>\r\n" );
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document.write( "        4a = 4 - 4a  ====>  4a + 4a = 4  ====>  8a = 4  ====>  a =  = .\r\n" );
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document.write( "Answer.  The radius of the small circle is  a = .\r\n" );
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