<p>I will post some math contest questions. these are mch harder than SAT questions, but if anyone can give some solutions, that would be great:</p>
<p>2 verticies of a square are A(log(15)5,o) and B(0,log(15)x) for x>1. the other 2 verticies are in the first quadrant. add the corrdinates of all four vertices and the sum is 8. compute x.</p>
<p>no one's interested?</p>
<p>lol i think you're looking for the AoPS forum</p>
<p>I second cory123.</p>
<p>But since you asked here:
clockwise
A(log(15)5, 0), B(0, log(15)x), C(log(15)5, 2log(15)x), D(2log(15)5, log(15)x).
4log(15)5 + 4log(15)x = 8
log(15)5x = 2
5x = 225
x=45.</p>
<p>These forums are not really about contests, at least not in a direct sense. :)</p>
<p>that only makes a rhombus, not a sq. but thanks, ok yeah i'll use the AOPS instead</p>
<p>FYI, nevermind, gcf you're right. you just messed up your coordinates but it didn't make a diff in the end</p>
<p>That's kind of a nifty question. Where did it come from?</p>
<p>A(log(15)5,0) and B(0,log(15)x) intersect the x and y axes, and if C and D must be in the first quadrant, then this is like a square sort of tipped up on its B corner. The distance between point B and (0,0) is the same distance between the horizontal distance between B and C, and the horizontal distance between B and A is the same as the vertical distance between B and C, so C is C(log(15)x,log(15)x+log(15)5). Then the vertical distance from the x axis to D is the same as the horizontal distance from (0,0) to A, and the horizontal distance from A to D is the same as the vertical distance from (0,0) to B. So D is (log(15)5+log(15)x,log(15)5). This is a lot easier to show with a picture, but I am too lazy to draw one so you'll just have to trust me. If anyone really cares THAT much I'll explain this all better but meh.</p>
<p>A(log(15)5, 0)
B(0, log(15)x)
C(log(15)x, log(15)x + log(15)5)
D(log(15)5+log(15)x, log(15)5)</p>
<p>Adding all the coordinates results in 8:
log(15)5 + log(15)x + log(15)x + log(15)x + log(15)5 + log(15)5 + log(15)x + log(15)5 = 8
4log(15)5 + 4log(15)x = 8
log(15)5 + log(15)x = 2
log(15)5x = 2
15^2 = 5x
225 = 5x
x = 45</p>
<p>came from ARML
since I've already started this thread, here's another one for anyone interested:
L is tangent to circle x^2 + y^2 = 1998 at T(a,b) in the first quadrant. If the interecepts of L and the origin form the verticies of a triangle whose area is 1998, compute the product ab.</p>
<p>As atonement for messing up the coordinates, notwithstanding my objection to this thread, heres the solution.</p>
<p>Naming conventions.
The origin O (duh).
OT = r
(r^2 = a^2 + b^2 = 1998).
Point T projects onto the X-axis at A, onto the Y-axis at B.
Y-intercept of L is C, X-intercept D.</p>
<p>From the similar triangles OTD and OAT:
OD/OT = OT/OA
OD/r = r/a
OD = r^2 /a.</p>
<p>From the similar triangles OTC and OBT:
OC/OT = OT/OB
OC/r = r/b
OC = r^2 /b.</p>
<p>Area of the triangle COD = (1/2) (OD) (OC) = 1998
(r^2 /a) (r^2/b) = (2)1998.
ab = (r^2)^2 / (2)1998
ab = 1998^2 / (2)1998
ab = 999.</p>
<p>To complete my atonement, heres the condensed version of the legendofmaxs solution (post #8). I am just as lazy to draw a picture. :p</p>
<p>Vertices in a clockwise order: A,B,C,D.
The origin O.
Point C projects onto the Y-axis at E.
Point D projects onto the X-axis at F.
OA = a, a = log(15)5.</p>
<h2>OB = b, b = log(15)x.</h2>
<p>Here goes.
The triangles BOA, CEB and AFD are congruent.
EB = OA = a, EC = OB = b, thus point C has the coordinates (b, a+b).
FD = OA = a, FA = OB = b, thus point D has the coordinates (a+b, a).
Adding all the coordinates of (a,0), (0,b), (b, a+b), and (a+b, a)
4a + 4b = 8
a + b = 2
log(15)5 + log(15)x = 2
log(15)5x = 2
5x = 15^2
x = 45.</p>
<p>Ahh I was gonna answer the circle one but gcf beat me to the punch -- I woulda literally done it the exact same way</p>
<p>Sorry!</p>
<p>As a consolation - a nifty corollary to that square question.</p>
<p>Let EC and FD intersect at G.
Any square inscribed in OEGF will have the same sum of its coordinates as OEGF.</p>
<p>thanks, gcf</p>
<p>Welcome!</p>
<p>It was a guilty pleasure to do those questions. :o</p>
<p>To phrase that corollary (the post above) better:
**
If a square vertex is in the origin and its two sides lie on the positive sides of the x- and y-axes, the sum of the vertices coordinates of any square inscribed in the big square is equal to the big square perimeter.**</p>
<p>Now I am truly done.</p>