# Math Problems for people who want a challenge

Nine math problems of varying difficulty with a separate pdf of solutions.

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Some problems of varying difficulty.pdf51.81 KB
Solutions- some problems of varying difficulty.pdf73.46 KB
jmerrante
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Joined: 11/09/2009
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I was working on #7 in the currently-posted (11/9/09) "problems of varying difficulty". The solution document says that the length of PS = 1/4*r*(1 + square root of 5). How can that be? The legs of the right triangle are r and 3r, so the hypotenuse PT would be r*(square root of 10) and PS is 1/4 of that: 1/4*r*(square root of 10). So the solution starts with an incorrect premise. What am I missing?

Also, at the end, it concludes that the measure of angle UPV is twice the measure of angle UPS, but S is NOT the midpoint of UV. The midpoint would be the point where line m meets PR, which is not S.

Can someone explain these seeming discrepancies?

David Wright
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Joined: 11/09/2009
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Problem 7

The legs of the right triangle are r and 2r since QR is TWICE the length of PQ. The hypotenuse PT is r*sqrt(5). Since line m is perpendicular to PR, PSU and PSV are right angles. The triangles PSU and PSV share a common leg and have hypotenuse a radius of the circle. So S is the midpoint of UV.

jmerrante
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Joined: 11/09/2009
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Problem 7 continued

But point S is not on PR -- it is on PT, 1/4 of the way between P and T. QT is length 3r, since RT = PQ = r. So the legs of the right triangle are r and 3r, with the hypotenuse PT being r*sqrt(10). Since S is not on PR, PSU and PSV are NOT right angles, and S is not the midpoint of UV. PS = 1/4*r*sqrt(10), but you still need to determine point (say) W, which is where line m intersects PR after passing thru point S. W is the midpoint of UV. It gets a bit more complicated from there...

Am I missing something about the placement of point S (on PT)?

jmerrante
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Joined: 11/09/2009
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correction

I see one problme with my calculations -- I was extending QR instead of extending PR -- back to the drawwing board!

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