Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest
Answer is here.
Ugly answer, but it works. Find the area of the whole triangle (Heron). Find the length of the altitude to BC. Find P(h) where the altitude falls on BC (Pythagoras). Apply the Pythagorean Theorem to find the square of the length of each Cevian. Add.
ReplyDeleteThere needs to be something nicer. Especially since the answer is so clean.
Jonathan
114.
ReplyDeleteOkay, this is how my husband and I did it. We know that a right triangle has sides proportional to: 1, 2, 3^1/2, where 2 is the hypotenuse. So using side AC=6 we can determine that ACP_2 is a right triangle giving q_2=3(3^1/2).
So now we know that using AP_2 we can use the pythagorean theorem to find the other values of q_n. (q_2)^2=27 So 27 + 1^2 = (q_1)^2 = (q_3)^2. Same procedure for q_4. Then you just square all the q_n values and add them up.
Erin, your husband seems to have assumed that ACB is 60º.
ReplyDeleteBut we do not know that ∠ACB is 60º. In fact, with some effort we can show that it cannot be (one way would be to use the law of sines to find the other two angles, which leads to a contradiction)
Without that 60, the proportion and the work that rests on it do not hold.
Jonathan
My method is similar to Jonathan's response, I guess. Assume A is (0, 0). First do law of cosine to "find" angle A (manually, which gives us A = cos-1(5/7)...) and then use Pyth Theorem to find sin(A) = 2*sqrt(6)/7. Then, use right-triangle trig to find the (x, y) values of C to be (30/7, 12*sqrt(6)/7). Then P1, P2, P3, P4 are just interpolated points along CB, which gives me their coordinates P1=(226/35, 12*Sqrt(6)/35), P2=(207/35, 24*Sqrt(6)/35), P3=(188/35, 36*Sqrt(6)/35), P4=(169/35, 48*Sqrt(6)/35).
ReplyDeleteThen finding q1^2 + q2^2 + ... is trivial albeit SUPER long. 150 is the sum.
The concept is not difficult once you wade through the red herrings. The arithmetic is long and nasty, since you have to do it all by hand. I used GeoGebra to plot the points / segment to check my work; the segment lengths check out (but I admit, I had to go back to check my work a few times to find the correct value for q1^2 + q2^2 + ... Apparently I suck at multiplying by hand?!)
Mimi,
ReplyDeleteto compare work (I included every step):
Early on I imagined the triangle with BC horizontal.
The semiperimeter is (5+6+7)/2 = 9.
The area is sqr[(9)(9-5)(9-6)(9-7)] = sqr(216).
The altitude can be found as (1/2)5h = sqr(216) so h = (2/5)sqr(216) = sqr(864/25).
Use CA = 6 or 900/25 and the altitude to find that P(h) is 6/5 from C.
Then Pythagoras using the altitude makes the qs: (1/5)^2 + h^2, (4/5)^2 + h^2, (9/5)^2 + h^2, and (13/5)^2 + h^2,
so the sum of the qs: 4h^2 + (1^2 + 4^2 + 9^2 + 14^2)/5^2.
Leave radicals unsimplified, leave fractions unreduced, until the end. It makes the arithmetic much less painful.
Nice! I like how our methods were actually pretty different, but they all led to Rome.
ReplyDeleteWoot!
ReplyDeleteJon's comment "leave unsimplified till the end" is usually the best way to go with these questions. Since there is no tech allowed, something HAS to cancel from the algebra or the question will ask for some construction (q1² + ...) that can be found easily in the 3-5 minutes available.
I think the easiest way do this is just to use the law of cosines throughout. (The fact that it wants you to find the lengths squared hinted me to this)
ReplyDeleteFirst find cos(C) by law of cosines
Cos(C)=[6^2-7^2-5^2]/(-2*7*5)=19/35
q_1^2=49+1 -2(7)(1)(19/35)
q_2^2=49+4 -2(7)(2)(19/35)
q_3^2=49+9 -2(7)(3)(19/35)
q_4^2=49+16-2(7)(4)(19/35)
Lining it up this way and using some quick mental math tricks makes the addition simpler:
SUM=4(49)+30-2(7)(10)(19/35)=150
This gets rid of all the tricky calculations