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Title: Math problem Post by: ns33 on January 21, 2004, 10:32:08 PM This math problem has been bugging me for almost a year and either I've been missing something for the past 12 months, or this problem is truly screwed over.
-edit- This is from the American Mathematics Competitions, Contest B, issued on Feb 26, 2003. The question is #22 Quote Let ABCD be a rhombus with AC = 16 and BD = 30. Let N be a point on AB, and let P and Q be the feet of the perpendiculars from N to AC and BD, respectively. Which of the following is closest to the minimum possible value of PQ? (A) 6.5; (B) 6.75; © 7; (D) 7.25; (E) 7.5 This is ALL the information there is. Please do not ask for more info, and do NOT make up information without being able to show it through a standard proof. Title: Math problem Post by: evilknight on January 22, 2004, 02:11:37 AM the answer is 42
Title: Math problem Post by: Rug on January 22, 2004, 10:34:21 AM Eeeek, thats evil. Maybe one of those unanswerable questions?
Title: Math problem Post by: opperdude on January 22, 2004, 10:52:50 AM well i'm just gonna start (dunno how to do the signs properly here but i'll try):
to make PQ as short as possible NP=NQ (rite?) APN~NQB AN=16/46 AB=16/30 BN BN=30/46 AB=1 7/8 AN AP=16/30 NQ =16/30 NP AN=16/30 BN NP=16/30 BQ i call the point where the diagonals cross point S: AS=8 (=CS) BS=15 (=DS) ok i'll move on later, but could sum1 tell me if this is right so far? cause i'd hate it if i assumed sumthing wrong... (and if sum1 knew a way to work out sum angles it would be easy :P ) Title: Math problem Post by: SS on January 22, 2004, 06:53:54 PM Can't be arsed with figuring out what opper is doing, but I'm 98.3274% certain that the solution requires the angle rules, so I'm going to go figure out if I can get anywhere using them.
Title: Math problem Post by: RipperRoo on January 22, 2004, 06:58:22 PM SOHCAHTOA?
Title: Math problem Post by: SS on January 22, 2004, 07:01:20 PM No, that's something different and only works for right-angled triangles. I was talking about the X/Z/F thing.
Although I've just spotted a reason why it might be a really easy problem... more in a sec. Title: Math problem Post by: opperdude on January 22, 2004, 07:06:55 PM me's waiting with tension...
and yeah i think u need to have some trick to figure out the angles or the length of AB, and after that you'll need what i did (i think) to get to an exact number Title: Math problem Post by: SS on January 22, 2004, 07:37:14 PM Gah, okay not as easy as I thought... this is tying my head in knots. :|
Title: Math problem Post by: ns33 on January 22, 2004, 09:05:53 PM The answer is 7.0, I figured it out yesterday. I've been assuming that its a parallelogram and not a rhombus, in which the diagonals would be perpendicular to each other. And since AB = BC = 17, it's easy to go from there. I originally thought I'd have to use law of cosines, but that wouldnt work if you arent given any other angles. I just forgot (for a year) the properties of a rhombus. :ph34r:
Title: Math problem Post by: SS on January 22, 2004, 09:11:16 PM Bah, I knew the diagonals were perpendicular, but none of the definitions mentioned it and your disclaimer made me believe you. Evil evil evil man! :angry:
Title: Math problem Post by: underruler on January 23, 2004, 05:01:31 AM Good thing I read all your posts or I would have been plagued too...I figured out that the perpendiculars were diagonal, but it's been quite some time since I've had some geometry problems. I'm so rusty. *creak creak*
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