dan's math@home - problem of the week - archives
Problem Archives page 13

1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+ . prob index

121 -- Gear Numbers
122 Power 2b Diffrnt
123 -- Tour de Ants !
124 Lattice Midpoint
125 Goats for Sheep
126 -- Longest Train!
127- Hex Floor Tiles
128 - Two or 3 Kids?
129 Grand Exponents
130 - Triang-U-Later!

 Problem #121 - Posted Monday, July 2, 2001 Gear Numbers ! (back to top) In the picture, assume the gears have 15, 16, and 17 teeth from left to right. Number each gear clockwise, 0-14 0-15, 0-16, with (0, 0, 0) at the top at noon. Assuming the gears take exactly one second per tooth, answer these: a) At what exact time will the gears first say (0, 0, 0) again? b) What's the first time the gears will say (1, 2, 3) ? c) What do the numbers say at 12:34 p.m.?

Solution: a) We are talking about Least Common Multiples here ; the LCM of 15, 16, and 17 is
N = 15 * 16 * 17 = 4080 , since there are no common factors. 1 hr = 60 min = 60(60 sec) = 3600 sec,
so 4080 seconds = 1 hr 480 sec = 1 hr 8 min past noon ; or 1:08 pm exactly.
b) Define a = b (mod m) to mean a and b have the same remainder when divided by m.
This is a setup known in Number Theory as the "Chinese Remainder Theorem" (CRT):
We seek a number x (mod 4080) with x = 1 (mod 15) , x = - 2 (mod 16) , x = 3 (mod 17).
(Since the second gear is moving in reverse, we need the minus sign.) One solution is
x = (1) * 16 * 17 * a + (-2) * 15 * 17 * b + (3) * 15 * 16 * c , where we have:
16*17*a = 1 mod 15 , 15*17*b = 1 mod 16 , and 15*16*c = 1 mod 17.
This makes the first term 1 mod 15 but 0 mod 16 and 17, the second term is 0 mod 15, etc.
Therefore 2a=1 mod 15 , a = 8 ; -b = 1 mod 16 , b = 15 ; 2c = 1 mod 17 ; c = 9.
x = 1*16*17*8 + (-2)*15*17*15 + 3*15*16*9 = 2176 - 7650 + 6480 = 1006 seconds.
Check: 1006 = 15*67 + 1 = 16*63 - 2 = 17*59 + 3 . It works ; this makes it
1006 sec = 16 min 46 sec past noon , or 12:16:46 p.m.
c) This is the much-simpler "CRT in reverse"; 34 min = 2040 sec. Divide by each number:
2040 = 0 mod 15 ; -2040 = 8 mod 16 ; 2040 = 0 mod 17 ; it says ' 0 , 8 , 0 '.

I like it when you tell me what software program or languange you used, if any (listed in purple)
Tim Poe . . . . . . . . 10 pts - (Excel) Good expl'n of LCM, remainders; 12:34 is halfway to 1:08.
Sue B . . . . . . . . . . . 7 pts - (Java) Nice recovery, you were still second with a correct solution.
Lisa Schechner . . . 5 pts - Got a good answer in under the wire! How'd you figure out b)?
Arthur Morris . . . .3 pts - (Matlab) Good, but ncluding the zeroes, that's mod 15, 16, and 17.
Hermen Jacobs . . . 3 pts - (Basic) 4080 sec is right but 68 min. And 23/30 not 23/40 in b).
Ludwig Deruyck . 2 pts - (Mathematica, Excel) All ok except middle wheel runs backwards.
Slanky . . . . . . . . . . 2 pts - (Calculator, pencil) Again, good but middle one runs backwards.

Problem #122 - Posted Thursday, July 12, 2001
From the set {0, 1, 2, 3, 4, 5, 6, 7}, pick three sequences of four distinct numbers (a, b, c, d)
so that placing parentheses in the expression a ^ b ^ c ^ d in all possible ways yields:
i) the fewest number of distinct values . . . ii) the greatest number of distinct values . . .
iii) the greatest number of prime values. Note: m^0 = 1 if m =/= 0; 0^p = 0 if p =/= 0; 0^0 is undefined.
There may be more than one sequence with the same number of values. Explain reasoning, minimize resubmissions.

Solution: There are five meanings with parentheses (no paren. is equiv. to #(5)):
(1) ((a^b)^c)^d = (a^b)^(c*d) = a^(b*c*d) . . . (2) (a^(b^c))^d = a^(d*(b^c))
(3) (a^b)^(c^d) = a^(b*(c^d)) . . . . (4) a^((b^c)^d) = a^(b^(c*d)) = a^(b^(cd)) ;
(5) a^(b^(c^d)) = a^(b^(c^d)) (this one's usually the largest)

(i) If a = 0, say 0^1^2^3, then all powers of 0 are 0, so there's only that one value.
(ii) Many choices give five values. For example, the five values for 3^4^5^6 are:
((3^4)^5)^6 = (3^20)^6 = 3^120 . . . . (3^(4^5))^6 = 3^(6*(4^5)) = 3^6144 ;
(3^4)^(5^6) = 3^(4*(5^6)) = 3^62500 . . . . 3^((4^5)^6) =3^(4^30) ~ 3^(1.153*10^18),
and finally 3^(4^(5^6)) = 3^(4^15625) ~ 3^(1.539*10^9407) (last 2 are approx.)
(iii) As pointed out by new contestant Guha Kaushik, the only way the answer is prime is
if a is prime and the exponent is 1. The choice 3^2^1^0 gives {1, 1, 9, 3, 9} one prime.

Ludwig Deruyck. . . . . 5 pts - I thought 4 values too but it's 5; primes are possible
Guha Kaushik (new) . . 4 pts - Welcome to dansmath.com; more than 3 values; good on iii).
Hermen Jacobs . . . . . . 3 pts - Greatest number of answers not greatest answer numerically...
Quasi-C. (new) . . . . . . . 3 pts - Sure, you can enter the contest, if I can contest your entry!

 Problem #123 - Posted Monday, July 23, 2001 Tour de Ants ! (back to top) Here are n ants who encounter some 'forks in the road.' A positive (whole) number of ants crawl along each path. RULES: (1) The net ant flow around a loop is zero, (2) The same number of ants go into a node as out. a) Find the smallest number, n, of ants that can do this. b) What nine values {a, b, . . . , i} will 'go with this flow'? Solution: Set up linear equations according to the above rules. Here's Guha's solution: The equations satisfying the given conditions are: (Dan's note: 6 nodes, 4 loops) 1) a+b=n . . . . 2) c+g=a . . . . 3) e +d=b . . . . 4) c+d=h+f . . . . 5) g+h+i=n 6) e+f=i . . . . 7) a+c-d-b=0 . . . . 8) g-h-c=0 . . . . 9) h-i-f=0 . . . . 10) d+f-e=0 from 4),6),9), we get f=(c+d+e)/3,i=(2e+c+d)/3,h=(2c+2d+e)/3 .... substituting f in (10), 4d+c-4e=0......(11)......from (3) and (11), 8d +c-4b=0.....(12) now, g=n-h-i =n-(c+d+e)... [putting the values of h & i] also, g-h-c=0 so, n-(c+d+e)-(2c+2d+e)/3-c=0 so..... 3n-(d+4b+8c)=0..........(13) from (1)&(7), n-2b+c-d=0....(14) .... solving (12),(13),(14), we get 33d=7n....The lowest integral value of n so that d is integral is 33 when d=7 which gives c=4, b=15, a=18, e=8, g=14, h=10, f=1, i=9. Dan's Note: Ok, this is the coolest thing: The flow diagram exactly corresponds to a tiling of a rectangle 33(=n) by 32(=a+g) with nine squares of sizes a, b, c, ..., i !! See problem #32 - Squares in a Box

Guha Kaushik. . . . . . 10 pts - Nice use of repeated substitution; thanks for your now-famous entry!
Quasi-C. . . . . . . . . . . . 6 pts - Your values are right; I'm tryina understand your excel matrix solution.
Phil Sayre. . . . . . . . . . 5 pts - Good to reduce to a triangular system of eqns; I like your 'sparse matrix'.
Arthur Morris . . . . . . 3 pts - The nodes are alright but there are some loop problems if n = 4 or 6.
Hermen Jacobs. . . . . . 3 pts - Fair try, but yes, it was a bit too simple if you don't use a+c=b+d, etc.
 Problem #124 - Posted Wednesday, August 1, 2001 Lattice Midpoint ? (Problem thanks to Mark Jaeger) (back to top) Below (in red) is a statement that may or may not be true. As we always say at Ohio State in the Arnold Ross Program, Prove or Disprove, and Salvage If Possible: "Given any nine lattice points in 3-space (each coordinate (a,b,c) is an integer), there is at least one pair of points whose midpoint is also a lattice point." Prove the statement is true or come up with a counterexample. If it's false, see if you can recover something that's true. Show reasoning. Solution: (From Hermen Jacobs) "The statement is true! A pair of two lattice points have an lattice- point as midpoint if every coordinate of the midpoint is an integer, also both coordinates of the 2 points are even (0) or odd(1) because (even+even)/2 = an integer and (odd+ odd) = also an integer, but (even + odd)/2 is not an integer. There are in 3-space 8 different possibilities of even (0) or odd(1) of the three coordinates: 1) 0,0,0 2) 0,0,1 3) 0,1,0 4) 0,1,1 5) 1,0,0 6) 1,0,1 7) 1,1,0 8) 1,1,1. If you have any nine lattice points in 3-space (one more than 8) the are also always two the same. These both points have also a lattice-point as midpoint. The statement is true!" Quasi-C asks: What's purple and commutes? Dan replies: An abelian grape!

Tim Poe . . . . . . . . . 10 pts - First four answers within 2 hours; yours was first fault-free offer.
Lisa Schechner . . . . 8 pts - Good explanation, bonus point for only being edged out by 1 minute!
Quasi-C. . . . . . . . . . 6 pts - Nice argt: Of 9, 5 must have same x-parity, 3 of 5 same y-p, 2 of 3 sm z.
Guha Kaushik . . . . 5 pts - Interesting use of probability: prove the prob of all same parities is > 0.
Hermen Jacobs. . . . 4 pts - Thanks for your continued support of my site! See above answer!.
Arthur Morris . . . . 4 pts - Early answer; 8th pt might not have same parity but another pair might.
Sam Wilson . . . . . . 3 pts - Your second answer filled in the logical gap of your first; good recovery!
Sue B . . . . . . . . . . . . 2 pts - The midpoint just needs to be a lattice pt, not nec. one of our nine pts.
Phil Sayre. . . . . . . . .1 pt - Just under the wire; same thing as Sue; any lattice pt not just blue ones.

Problem #125 - Posted Sunday, August 12, 2001 ... nearing the end of the 2000-01 contest!
Marlee and Charlie decide to sell all their sheep and go into goat herding(!) They get
as many dollars per sheep as they had sheep, and they buy as many \$10 goats as they
can with the money. This leaves them with a few dollars, with which they buy a rabbit.
They now have an even number of animals which they split evenly; Charlie has all goats.
How much money should he give to Marlee to even up the value of their parts?

Solution: If they had n sheep then the got n^2 dollars for them. The number of goats they get is the
number of tens in n^2. There's a quirk in the list of squares: if the tens digit is odd, the units digit is a 6.
(Checking the first 25 squares is sufficient.) So with an odd number of goats, the one rabbit costs \$6,
the odd goat was \$10. So to make it even, Charlie gives Marlee (10 - 6)/2 = \$2 to make them even.

Dan's note: Special kudos go to new contestant Steve Lawrie for a great answer with a sense of
discovery, and a quote from Charles Darwin mixed in; I love the quote but it sounds frustrating...
"A mathematician is a blind man in a dark room looking for a black cat that isn't there."

Ludwig Deruyck. . .10 pts - Good use of Excel and logic combined! Rabbit = \$6 if #goats = odd.
Arthur Morris . . . . 7 pts - Noticed 'lack of info' was the key ; right that #sheep ends in 4 or 6.
Sudipta Das . . . . . . 6 pts - Nice : x^2 = 10(2y-1)+z gives x ending in 4 or 6, so z=6 proves it.
Lisa Schechner. . . . 5 pts - Excellent & thorough: rabbit=\$6 even though #sheep or goats unknown.
Hermen Jacobs. . . . 4 pts - That's right, give \$2, half the difference. Good system of equations!
Tim Poe. . . . . . . . . . 4 pts - If x=#sheep then (x^2 - (x^2 mod 10))/10 = #goats ; good follow-up!
Sue B . . . . . . . . . . . . 3 pts - Wow, a ruthless nested looping Java program trained on goats & sheep!
Quasi-C. . . . . . . . . . 3 pts - Yes you are using quadratic residues mod 10, meaning 'last digit' <;-}
Guha Kaushik . . . . 3 pts - I agree that s^2 - d = 10g and that the rabbit costs \$6, so give \$2 to M.
Steve Lawrie (new). . 3 pts - Welcome! Yes, odd tens => six ones, but give \$2 to even up. Great quote!
Phil Sayre. . . . . . . . .2 pts - Your equations were there and could be solved, then C gives M half diff.
Basem Chaikhouni. (new). 1 pt - Good start, welcome to the contest! Read the surprising solution.

Problem #126 - Posted Saturday, August 25, 2001 ... last problem of the 2000-01 contest!
The Longest Train (Clues for values of n, m, and p are given below.) (back to top)
The world's longest train is n miles long, and takes m minutes to pass a certain point.
A train robber on horseback can ride alongside from the rear of the moving train to
the front and back to the rear in p minutes. How fast is the horse?
Clues: 2 < n < m < p ; n+m+p = 40 ; n is prime ; p is a mult of 10 ; m is a mult of n.

Solution: First of all, we find n, m, and p. We must have n = 5 and so 5 + m = 40 - p is a mult of 10.
This gives m = 15 and p = 20. Now the train goes 5 miles in 15 min which is 20 mph.
Now from Tim Poe: "Relative to the train, the horse travels N = 5 miles at a speed of H-20mph from
caboose to engine plus 5 miles at a speed of H+20 mph from engine to caboose.
The total distance is traveled in a total time of 20 minutes = 1/3 hour.
1/3 = 5/(H-20) + 5/(H+20) ==> 1/3(H-20)(H+20) = 5(H+20) + 5(H-20) = 10H
(H-20)(H+20) = 30H ==> H^2 - 30H - 400 = 0 ==> (H+10)(H-40) = 0 ==> H is -10 or 40."
The horse runs at 40 mph. (Does H = -10 make sense in some weird time-reversal sense?)

Lisa Schechner . . . . 10 pts - Ok, I trust you with the m, n, and p. You drive the car, I'll ride a horse.
Sudipta Das . . . . . . . 7 pts - Interesting ; s = speed of horse, Time = 2 s n / (s^2 - (n/m)^2). Yes.
Arthur Morris . . . . . 5 pts - Trial and error is often the best way to whittle down a known list.
Tim Poe . . . . . . . . . . 4 pts - Thorough expl of m,n,p and use of the above quadratic eqn. Thanks.
Steve Lawrie . . . . . . 3 pts - The clues were put there to help; thanks. I mean you're welcome.
Phil Sayre . . . . . . . . . 3 pts - Another case of process of successive elimination of possibilities.
Quasi-C . . . . . . . . . . 3 pts - No quadratic residues this time, just quad eqn. h = 2/3 whats per min?
Derek Chin (new) . . . 2 pts - Welcome to the contest; good setup of equation, slightly slow horse.
Hermen Jacobs . . . . 2 pts - Kudos for the expl'n of the m,n,p; your horse is the speed of the train?
Payman Parang (new) 2 pts - Thanks for entering; speed of the train isn't same rel to horse as to ground
Mike Fuqua (new) . . . 1 pt - Good try; with some more steps I can see how you got your answer.

 Problem #127 - Posted Thursday, September 6, 2001 first problem of 2001-02 contest! Hexagonal Floor Tiles (back to top) The hexagonal tiles on my large bathroom floorare 2 inches across from the middle of one side to the middle of the opposite side. They are separated by regions of white cement that are all 1/8 inch wide. What percentage of the floor covering is cement? Answer exactly and also (if necessary) to the nearest 1/100%. Show reasoning, keep resubmissions to a minimum. Solution: Compare the colored hexagons with a hexagon that contains it and bisects the thin strips of cement; that one has a height of 2 + 2(1/16) = 2 1/8 inches. The ratio of the areas will be the square of (2 1/8)/(2) = 17/16 ; or 289 / 256. The cement occupies (289 - 256) / 289 or 33/289 of the total (colored hex + cement); this is (3300/289)% ~=~ 11.42% of the floor.

Lisa Schechner . . . . . 10 pts - Similar hexagons, good. A bit off in your 3rd dec pl but same to 100th%.
Arthur Morris . . . . . . 7 pts - Yes, 'large' means ignore the edges, and as you say, less is more!
Hermen Jacobs . . . . . 5 pts - The height and base are both multip by 17/16, but your ans was ok.
Phil Sayre. . . . . . . . . . 4 pts - I like the (1/1.0625)^2 way of expressing it; the 'drill-bit' decimals!
Quasi-C. . . . . . . . . . . . 4 pts - By using narrow rooms, QC also proves 5.9% < grout < 20.7% ;-}
Yehuda Levy (new) . . . 3 pts - Welcome to the contest; square 1 1/16, yes; glad you like my site!
Steve Lawrie . . . . . . . 3 pts - Dissected hex & grout into rectangles and triangles; that also worked!
Tim Poe . . . . . . . . . . . 3 pts - Good geometry to get 2 / sqrt[3] but the area is prop to length squared.
Navdeesh Sidhu (new) 2 pts - Thanks for entering; looks like you used a sector of a circle instead of hex.
Payman Parang. . . . . 2 pts - The 6 / sqrt[3] was correct for one hex, then count in corners too.
Basem Chaikhouni . 2 pts - Also the right hex area but not sure where the 4 * grout= 80 comes from.

Problem #128 - Posted Saturday, September 15, 2001
"It's been four years since I saw you," said Martha. "How old are your two kids now?"
"Gee, it has been a while; I have three kids now!" replied Suzanne. "If you multiply
their (integer) ages it's 2/3 of a gross, but if you add them up you get your present age."
Martha said, "That still doesn't pin it down for me!" Suzanne winked, "Yes it does;
think about it!" How old are the three kids now, and why?

Solution: This sure reminds me of the Census Taker Problem, found elsewhere in my archives!
First we see that 2/3 gross = (2/3)(144) = 96 ; look for a*b*c = 96 ; and compare the sums.
We see there are a dozen possible ways including the improbable 1 + 1 + 96 = 98 ; there's also
1 + 2 + 48 = 51 ; 1 + 3 + 32 = 36 ; 1 + 4 + 24 = 29 ; 1 + 6 + 16 = 23 ; 1 + 8 + 12 = 21 ;
2 + 2 + 24 = 28 ; 2 + 3 + 16 = 21 ; 2+4+12 = 18 ; 2+6+8 = 16 ; 3+4+8 = 15 ; 4+4+6 = 14.
Then Martha must be 21 since she was still unsure. But if the kids were 2, 3, and 16, four years
ago there woulda been only one kid, not two as stated. So they're 1, 8, and 12 years old.
Dan's note: Quasi-C suggests if second kid is 3 yrs 10 mos then 4 yrs ago Suzanne was 7 mos pregnant,
Martha woulda noticed, and hence the wink; ok if ages 2, 3, 16 Martha would assume there were 2 kids now.

Sudipta Das . . . . . . . 10 pts - Good concise answer but not lacking any key details!
Lisa Schechner . . . . . 7 pts - Right, the age has to be the one that has two ways of summing.
Quasi-C. . . . . . . . . . . . 5 pts - I like it when you stand up for your (original) answer! (see above)
Arthur Morris . . . . . . 4 pts - Good answer, and yer right; we're ignoring deaths, adoptions, etc!
Zahi Yeitelman (new) . 3 pts - Welcome to the contest! I also forgot the case 4*4*6 but same answer!
Steve Lawrie . . . . . . . 3 pts - Interesting you eliminated all a < 4 first, then there was no ambiguity!
Tim Poe . . . . . . . . . . . 2 pts - Ok to interpret the wink but not nec Martha = '29'; also 2+3+16=21
Hermen Jacobs . . . . . 2 pts - Thank you for your kind comments, Wide range of ages possible!
Dan Baczkowski.(new) 1 pt - Welcome to dansmath - remember 'reasonable' not required for ages.
Mike Fuqua . . . . . . . . 1 pt - Right; x y z = 96 ; and there are a few possible values of x + y + z.
Navdeesh Sidhu . . . . . 1 pt - Good to list all possible ages first, what about a 96-yr-old ''kid''?
Cindy Tang.(new). . . . . 1 pt - Thanks for entering! You don't have to be 'mature' to talk to a mom.
Anirban Bhattacharyya.(new) 1 pt - Good to notice if x < 4 then yz > 24. See you on a future problem?
Sukhmine Bains (new) . 1 pt - Welcome! Got in just under the wire. A gross equals a dozen dozen

Problem #129 - Posted Tuesday, September 25, 2001
Consider these two expressions raised to the same power:
(a) (1 + x^2 -- x^3)^1000 . vs . (b) (1 -- x^2 + x^3)^1000
When expanded, which one has the larger coefficient on the x^24 term, and why?

Solution: We can substitute (-x) for x to see what happens;
a(-x) = (1 + (-x)^2 -- (-x)^3)^1000 = (1 + x^2 + x^3)^1000 , while
b(-x) = (1 -- (-x)^2 + (-x)^3)^1000 = (1 -- x^2 -- x^3)^1000 .
Since (-x)^24 = x^24, the coefficient of x^24 is the same in a(x) as in a(-x) , likewise with b(x). . . .
But the terms of (a) are all positive, while the terms of (b) are mixed pos and neg, and add up to less.
So the coeff of every even power, incl. x^24, of x must be smaller in b(x) and larger in a(x).
Dan's Note: Quasi-C & Tim Poe calculated the actual approx coeffs: a(x): 3.31*10^27, b(x): 7.05*10^26.

Arthur Morris . . . . . 10 pts - Listed the five possible ways of getting x^24; first has all +.
Hermen Jacobs . . . . . 7 pts - I like the "1^p(+-x^2)^q (-+x^3)^r , p+q+r=1000" method.
Quasi-C. . . . . . . . . . . . 5 pts - I'm pretty sure I follow your J=994,1000-J=6,(x^3)^6 = x^18.
Tim Poe . . . . . . . . . . . 4 pts - 5 ways: (x^2)^12, (x^2)^9(x^3)^2, . . . , (x^3)^8. Mostly +!
Steve Lawrie . . . . . . . 3 pts - I didn't know there was a "usual trinomial expansion", now I do!
Lisa Schechner . . . . . 3 pts - Said both are (1+-z)^1000, where z = (1 - x)x^2; use binomial!
Sudipta Das . . . . . . . . 3 pts - Unique method giving exact answers in terms of combinations.
Mehdi Ansari (new) . . 2 pts - Welcome to my contest! Smaller exponents don't tell the whole story!
Cristina Cancio (new). 2 pts - Hello, thanks for entering; good intuition on more x^2 canceling out.
Zahi Yeitelman . . . . . 2 pts - Correct about the (-x^3)^9 giving neg coeff; a bit off on x^976
Erudite1A (new) . . . . . 1 pt - Welcome, just under the wire. Power 12 gives a good idea but not same coeff.
Phil Sayre. . . . . . . . . . 1 pt - Really just under the wire, b4 I had soln up. I like the double sum.

Problem #130 - Posted Thursday, October 4, 2001
I drew four triangles with integer sides. Two were right, and two were not. The right ones had area
numerically equal to their perimeter, and the others' areas were two-thirds of their perimeters.
What were the sides, perimeters, and areas of my four triangles? (Hint: One of the triangles was rather thin!)

Solution: The right triangles have perimeter P = a + b + c = a + b +(a^2 + b^2), and area A = a b / 2.
getting ab - 2a - 2b = 2(a^2 + b^2) or (ab - 2a - 2b)^2 = 4(a^2 + b^2). Expand, collect, and presto...
This gives the two answers (a,b,c) = (6,8,10) , A = P = 24 ; and (5,12,13) , A = P = 30. (You do the math!)

The others have perimeter P = a + b + c and by Heron's formula, area A = [s(s-a)(s-b)(s-c)] ; s = P / 2.
An exhaustive check up to a=b=c=100 reveals (5,5,8) , A = 12 , P = 18 ; and (3,25,26) , A = 36 , P = 54.
The thin one is 3, 25, 26 with an angle of about 6.36 degrees, found by the law of cosines (link to my lessons).

The extension to Heron's formula was first sent in by Tim Poe, who said the area of quad abcd is
1 ________________________________________________________
--- /(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d) - 16 abcd cos(q)^2
4
. . . . . . . . . . . . . . . .. . .. . . . . . . . . . . . . . . . . . .as determined by the hindu mathematician Brahmagupta.
(Dan's Note - - Good job Tim! As you didn't clarify what q was, I also award a point to Lisa Schechner:
Heron's formula for quadrilaterals: Introduced by an indian mathematician Brahmagupta and is given by:
Area = sqrt( (s-a)(s-b)(s-c)(s-d)-abcd*cos((Z+X)/2))
(Dan's note: is this indpt of choice of pair Z,X?)
where s=(a+b+c+d)/2, and Z,X are any 2 opposing inside angles in the quadrilateral.
A special case is a cyclical quadrilateral (one that sits on a circle) for which the formula is reduced to Area =
sqrt((s-a)(s-b)(s-c)(s-d)). and, of course, Heron's formula is a special case of a cyclical quadrilateral with d=0.
(Alex Chen says this formula is attributed to Heron of Alexandria but can be traced back to Archimedes.)

Tim Poe . . . . . . . . . . . 11 pts - The winner and a bonus research point! Go Visual Basic!
Arthur Morris . . . . . 7 pts - Also provided (6,25,29), (9,10,17), and (7,15,20) in which A=P.
Lisa Schechner . . . . . 6 pts - (incl bonus) Heronian Triangles is a term I hadn't heard; thanks!
Sudipta Das . . . . . . . 4 pts - You were next in line; a good solution, but light on the details.
Payman Parang. . . . . 3 pts - Got two of the four but you & Derek pointed out unclear wording.
Zahi Yeitelman . . . . . 3 pts - Heron's equiv to (P-a)(P-b)(P-c) = 16P/9. I did not know that!
Quasi-C . . . . . . . . . . . 2 pts - No, a (1,3,4) has A = 0 but P=/=0. I liked your other three, tho.
Steve Lawrie . . . . . . . 2 pts - Found 3 of the 4 without benefit of Heron; good job!
Derek Chin . . . . . . . . 2 pts - Right to wonder why I'd have 2 of each; I meant all 4 different.
Phil Sayre. . . . . . . . . . 2 pts - x=n(m^2+h^2), y=m(n^2+h^2), z=(m+n)(mn-h^2) gives us a
. . . . . . . . . .triangle (x,y,z) with A = hmn and P = 2mn(m+n) (So make h=2(m+n) or 4(m+n)/3.)
Sabina Paradi (new). . 1 pt - Thanks for entering! 6,8,10 ok, and so is .5 ab sin t = 2/3 P.
Alex Chan (new) . . . . . 1 pt - Hello, got 5,12,13, the 6,8,12 not 'right', thanks for Archimedes!
Cristina Cancio. . . . . 1 pt - The two right tri's were spozed 2B diff but 5,12,13 works.
Adel Farahmand(new) 1 pt - Hi Adel; not sure what a^2+b^2+c^2 measures but it's interesting.

 THANKS to all of you who have entered, or even just clicked and looked. My site is now in its fifth season - OVER 27,000 HITS so far! (Not factorial.) Help it grow by telling your friends, teachers, and family about it. YOU CAN ALWAYS FIND ME AT dansmath.com - Dan the Man Bach - 3*23*29 A.D.

Problem Archives Index

Probs & answers . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90
Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90

Probs & answers . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+
Problems only . . . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+

Browse the complete problem list, check out the weekly leader board,
or go back and work on this week's problem!