Question

Easy, simple way to do this? The perimeter of right triangle RST is equal to the perimeter of isosceles triangle XYZ. The lengths of the legs of the right triangle are 6 and 8. If the length on each side of the isosceles is an integer, what is the greatest possible length for one of the sides of the isosceles triangle XYZ?

A) 10
B) 11
C)14
D)16
E)22

Answer 1

@phi could u please help in this? the easiset way to solve? i know how to solve it, but it is way too long. this is from the PSAT.

Answer 2

first, we should recognize that the right triangle is a scaled version of a 3,4,5
(because 6 and 8 are 2 times 3 and 4)
perimeter of 3,4,5 is 3+4+5= 12. 2 times that is 24 for our right triangle

Answer 3

right.

Answer 4

all this time in my head while iwas solving this problem i was thinking that the two sides of a triangle should always be greater than the third. So i relied on mostly trial and error reasoning. but, i think this kind of method i use is too time-consuming

Answer 5

next, we can say
2x+y = 24

where x and y are the legs of the isosceles triangle
y = 24 - 2x

now the tricky part. if x =1 we get y= 22 . but if we try to draw that triangle we get

which does not work. we need to use the idea that two legs of a triangle add up to a longer length than the 3rd leg

Answer 6

so we need 2x> y (if 2x=y we do not get a triangle)
but we know y= 24-2x so we replace y to find
2x > 24 - 2x
add 2x to both sides
4x > 24
x> 6
so x has to be bigger than 6
x=7 is the next number bigger than 6


if x=7 then y = 24 - 2*7= 24-14=10

Answer 7

how does this also ensure that x+y from the isosceles triangle is greater then the other x(also from the isosceles)?

Answer 8

x+ y will always be > x (assuming y is positive)

Answer 9

oh because y cannot be 0. 0 is not an integer as well correct?

Answer 10

y is the length of one of the sides. 0 sounds just a bit too short, doesn't it ?

Answer 11

yes. i would agree on that. otherwise it would look just like a line... so with these kind of problems we have to assume y is an integer greater than 0? because it demands that XYZ is isosceles...

Answer 12

0 is an integer (as are negative numbers)... it is unstated they mean positive integers, because we are dealing with lengths of the sides of a 3 sided object.

Answer 13

oh ok. thank you so much! that was indeed a faster method than mine!!

Answer 14

the only problem is that it took me forever (relatively speaking) to get the logic. If I had to figure it on the fly it might take longer than trial and error...
on the other hand, having seen this problem, I would probably remember how to solve it the quick way....

Answer 15

ok.

Answer 16

the PSAT probably expected the student to solve this based on trial and error or sthing then... but in case the PSAT introduces a similar problem on the actual, i wanted to be prepared for it in advance and finish the math section within the 25 min time limit.

Answer 17

looking at it again, trial and error is the best strategy...
because if we have 11,11,2 that is a valid triangle. 11 is the longest side
12,12,0 won't work

so my logic up above ( x> 6) only found the longest y side = 10
but x can be longer than 10...

Answer 18

this problem is really complex. yeah i just looked at the answers and the correct answer was b. your second thinking is right.

Answer 19

each of the letter options can serve a either x or y of the isosceles, so we have to test it-2 times for each option- for a total of 10 times.

Answer 20

yes, I am trying to figure out the intuition you need so you can rule out some or most of the choices.

Answer 21

how about this idea... if we flatten the triangle we get


there is no way to get a side as long as 12 (isosceles or not)
so the choices are 10 or 11

Answer 22

that makes sense.

Answer 23

so the idea would be to test 11 first... we are down to testing if x can be 11 or y can be 11

Answer 24

x can be 11 24-11=13 so y CANNOT be 11. its a matter seeing whether the difference yields an odd or even number i think. if odd, we can rule out that choice since x must be integers.

Answer 25

actuallly i was wrong..

Answer 26

so since x can be 11 choice b has greater side length.

Answer 27

yes, the odd/even idea occurred to me... but noticing it on the test might not happen.. until after you check if y=11 works. But if you realized immediately that y cannot be odd (because that leaves an odd number to be divided by 2 for x), you are down to checking if x=11 works

Answer 28

In all of this, I am defining x to be the length of the = sides, and y is the base

Answer 29

yeah ok. i understand that part..really complex stuff.. this problem must take the most amount of time and patience..plus we have to compare the lengths. ok thank you for your help. this was really helpful!