**dan's math@home - problem of the week - archives****Problem Archives**page 13**with Answers and Winners.**For Problems Only, click here.**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 index121 -- 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 '**.

**WINNERS - Problem 121 .**(back to top) . leader board .- 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
- Power To Be Different ! (back to top)
- 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.**

**WINNERS - Problem 122 .**(back to top) . leader board .**Ludwig Deruyck**.**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**

**WINNERS - Problem 123 .**(back to top) . leader board .**Guha Kaushik**.**Quasi-C**. . . . . . . . . . . . 6 pts - Your values are right; I'm tryina understand your excel matrix solution.**Phil Sayre**. . . . . .**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!

**WINNERS - Problem 124 .**(back to top) . leader board .**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**.**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**. . . . . .- Problem #125 - Posted Sunday, August 12, 2001 ... nearing the end of the 2000-01 contest!
- Goats for Sheep! (back to top)
- 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."**

**WINNERS - Problem 125 .**(back to top) . leader board .**Ludwig Deruyck**.**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**.**Steve Lawrie**(new). . 3 pts - Welcome! Yes, odd tens => six ones, but give $2 to even up. Great quote!**Phil Sayre**. . . . . .**Basem Chaikhouni**.- 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?)

**WINNERS - Problem 126 .**(back to top) . leader board .**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**. . . . . .**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.

**WINNERS - Problem 127 .**(back to top) . leader board . (First problem of new contest!)**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**. . . . . .**Quasi-C**.**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
- Two or Three Kids... (back to top)
- "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.
**WINNERS - Problem 128 .**(back to top) . leader board . (Second problem of new contest)**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**.**Arthur Morris**. . . . . . 4 pts - Good answer, and yer right; we're ignoring deaths, adoptions, etc!**Zahi Yeitelman**(new)**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
- Grand Exponents! (back to top)
- 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.**WINNERS - Problem 129 .**(back to top) . leader board .**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**.**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**.**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**. . . . . .- Problem #130 - Posted Thursday, October 4, 2001
- Triang-U-Later ! (back to top)
- 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.)

**WINNERS - Problem 130 .**(back to top) . leader board .**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**.**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**.**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**. . . . . .- . . . . . . . . . .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 **- Dan the Man Bach**- 3*23*29 A.D.

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