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Author Topic: Math problem  (Read 6339 times)
ns33
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« 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.

 
« Last Edit: January 22, 2004, 09:06:11 PM by ns33 » Logged
evilknight
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« Reply #1 on: January 22, 2004, 02:11:37 AM »

the answer is 42
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« Reply #2 on: January 22, 2004, 10:34:21 AM »

Eeeek, thats evil. Maybe one of those unanswerable questions?
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opperdude
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« Reply #3 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 Tongue )
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SS
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« Reply #4 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.
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Peter 'SpectralShadows' Boughton,
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« Reply #5 on: January 22, 2004, 06:58:22 PM »

SOHCAHTOA?
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« Reply #6 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.
« Last Edit: January 22, 2004, 07:03:44 PM by SS » Logged

Peter 'SpectralShadows' Boughton,
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« Reply #7 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
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« Reply #8 on: January 22, 2004, 07:37:14 PM »

Gah, okay not as easy as I thought... this is tying my head in knots. :|
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Peter 'SpectralShadows' Boughton,
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ns33
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« Reply #9 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:  
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SS
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« Reply #10 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:
« Last Edit: January 22, 2004, 09:11:30 PM by SS » Logged

Peter 'SpectralShadows' Boughton,
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Till shade is gone, till water is gone, into the Shadow with teeth bared, screaming
defiance with the last breath, to spit in the Sightblinder's eye on the Last Day.
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« Reply #11 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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