<p>lol.....where have i seen this que before....i can't remember....amyb somewhere.....</p>
<p>Another lunchtime (for me) SAT math question. This one is pretty hard. I would consider you well-prepared for SAT I geometry if you can get it (without using a ruler!)...</p>
<p>it's about 2.11 ish for Fignewtons question</p>
<p>i got 4.24</p>
<p>nvm, thought there was a right triangle</p>
<p>No, it's 3.6</p>
<p>its 3.6,i am sure</p>
<p>if you know your high school geometry well,then u must know the relation between a tangent and a circle-disecting line and for this diagram the relation is AD^2=BD.PD</p>
<p>so---AD^2=BD(BD-PB)....OR, AD^2=BD^2-BD.PB....OR,PB=(BD^2-AD^2)/BD....THAT MEANS---- PB=(100-64)/10..(SINCE BD=10 AND AD=8)....PB=3.6</p>
<p>BY THE WAY --- is this a SAT l math problem?it's ALMOST impossible to find out if you don't know the theorem,unless u know trigonometry like **It'sGr82BeAGator**</p>
<p>I guess I'll explain one way to do it.
BC is a radius. So is PC, both of which have length 3. Together, BC, PC, and PB form a triangle. We know two of it's sides are the same, PC and BC, now we just need to find PB.
Looking at triangle BAD, it's a right triangle with sides 6 and 8, making the hypotenuse 10. Since we know 3 sides and one angle, we can certainly find both unknown angles. We find the measure of angle ABD, as that's the one that'll help most in this problem. so cos B=6/10, making angle B=53.13 degrees or something like that. Since triangle BPC is isosceles, and angle B is opposite one of the radii we know one of the angles must be 53.13 degrees as well. Then just find the last angle and plug into the law of sines to get the answer</p>
<p>^wow gator.....u made me edit my post</p>
<p>Another way to do this is to recognize that PB and PA form right angles, so we can use that too. We know we have a 6-8-10 right triangle in triangle BAD, and since PB and PA form a right angle, triangle PAB must be a right triangle. And by the complementary rule, PD and PA must form right angles as well. so PAD is a right triangle too.
From this we can set up the following equations
PB + PD = 10
PB^2 + PA^2 = AB^2 = 36
PD^2 + PA^2 = AD^2 = 64
To eliminate one variable, we subtract the bottom two equations, leaving us with PD^2 - PB^2 = 28
Now we're left with
PB + PD = 10
PD^2 - PB^2 = 28</p>
<p>Easily solvable by substitution
PD = 10 - PB
(10 - PB)^2 - PB^2 = 28
100 - 20PB + PB^2 - PB ^2 = 28
100 - 20PB = 28
72 = 20PB
3.6 = PB</p>
<p>^i didn't go this way coz it was time consuming.......besides some of us might get confused how PB and PA form a right angle....(coz AB is the diameter of the circle and P is a point on the circumference)
but nontheless,a great method</p>
<p>i thought trig functions could only be used with right angles?</p>
<p>edit: triangles*</p>
<p>ok, i got it.
i used an extra step at the end in gator's first solution.
angle C to midpoint of B and P makes two congruent right triangles.
then cos 53.13 = a/3
a=1.8
2a=BP=3.6</p>
<p>WOW
i copied the problem down incorrectly onto my piece of paper
rather than a 6-8-10 triangle i had something else really off wow hahahah
i dont even know how i did that
but yes, with the proper triangle, 3.6 should be the answer</p>
<p>nope october....trigonometry has much more depth...... :(</p>
<p>Sorry, i should of been more clear. I meant the trig functions such as sine, cosine, and tangent. to use them i thought we needed a right triangle.</p>
<p>I also found this definition - trigonometric functions are commonly defined as ratios of two sides of a right triangle containing the angle.</p>
<p>Since this is an SAT-level problem, trig is not required (of course, if it worked for you, who am I to say?).</p>
<p>Did anyone notice two similar triangles? Namely, ABP and ABD.</p>
<p>lol - I haven't done geometry in ages, so I was initially stumped.</p>
<p>I looked through all these overly-convoluted answer methods, until I remembered a bit about similar triangles. ABP is similar to ADP - the proportion BP/6=6/10 can be set up. Then, BP = 36/10 = 3.6.</p>
<p>Much quicker ;)</p>
<p>Wow, archonx just owned everyone in this thread. lol</p>
<p>Regardless, thanks for the demos and Gator's and Gluttony's solutions are pretty clever too.</p>