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A and B are 2 points 7cm apart. Construct the positions of point P which is 4.2cm from A and B 5.6cm from B. How many possible positions for P are there. Measure the distance between them.

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I need you to change the title it is important right now. –  Asaf Karagila Nov 29 '12 at 15:06
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Give me more time, please. –  B. S. Nov 29 '12 at 15:09
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Will the answer be important later as well? –  Graphth Nov 29 '12 at 15:10
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OK, we've had our fun. I've edited in a more informative, if less amusing, title. –  Gerry Myerson Nov 30 '12 at 2:01
    
@Gerry: Did you do it right now? It was important. :-) –  Asaf Karagila Nov 30 '12 at 14:44
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3 Answers

Note that the distance of each point P to A is 3/5ths of the distance from B to A, and also the distance of P to B is 4/5ths of the distance from B to A. So we are dealing with two points P that are vertices opposite the common hypotenuse of mirror-image right triangles.

This is easily constructed with (collapsing) compass and (unmarked) straightedge. Start by dividing AB into five equal intervals. Then use the respective points 3/5th from A and 4/5ths from B to draw a pair of (unequal) circles about A and B, giving two points of intersection as the possible solutions P.

The line segment connecting the two possible solutions P is then perpendicular to the hypotenuse AB. Applying a well-known result about the altitude to the hypotenuse of a right triangle, we can work out that the distance between the two solutions P is 24/25ths of the distance from B to A. Using that AB has length 7cm, the absolute distance between the two points P is then easily calculated.

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Hint: you might as well take $A$ at the origin and $B$ at $(0,7)$. Do you know the formula for a circle with given center and radius?

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No, he doesn't. Give him the formula right now –  Adam Rubinson Nov 29 '12 at 15:25
    
YEAH ITS PIE R SQUARED –  Sumaiyah Nov 29 '12 at 15:34
    
"YEAH..." Desperate and a smart-aleck too. –  Dilip Sarwate Nov 29 '12 at 15:42
    
The word "construct" in the question suggests a geomtric (compass & straightedge) approach, rather than a Cartesian coordinate solution/formula. –  hardmath Nov 29 '12 at 15:51
    
@Sumaiyah: No, it is not. Not even if you know the constant is $\pi$ instead of pie and stop yelling. I was trying to see what you do know and what class it is in, so I could provide a useful hint, but I have lost interest. –  Ross Millikan Nov 29 '12 at 16:13
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To be of 4.2cm distance from $A$ means to be on the circle around $A$ with radius 4.2cm.

Now draw those circles, and find all $P$ points that satisfy the requirement.

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