Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

Sure, contents list would be helpful!

Quick last question, if ACH, BHD and HAB are congruent (SAA) how is it that AHB = 90?

Thanks, but i don't really understand how you got AB = HD = CH or how you got ACP + BDP = 90.

**championmathgirl**- Replies: 39

Two circles are externally tangent at point P. Segment CPD is parallel to common external tangent AB. Prove that the distance between the midpoints of AB and CD, is AB/2.

only without the thin drawn in lines. I couldn't find a better photo.

Is r=(1+sqrt3)/2? And is Cos 36 = 1/2+1/2*r, Cos 72=1/(2+2r)

I checked it, and it didn't seem to work out. Where is it that I went wrong?

Erm... I'm kinda stuck. I got to part (c) and well I got the equation r^2+xr-1=0. Even using the quadratic or completing the square, i get stuck with a square root with a lot of variables.

I want to check to make sure the equation is right, and what i got for r is also right. Then I need a bit of help with the rest.

Super thanks!!

**championmathgirl**- Replies: 21

In triangle ABC, angle A = 36 degrees and angle B = angle C = 72 degrees. Let BD be the angle bisector of angle ABC.

(a) Prove that BC = BD = AD.

(b) Let x = BC and let y = CD. Using similar triangles ABC and BCD, write an equation relating x and y.

(c) Write the equation from Part (b) in terms of r=y/x and find r.

(d) Compute cos 36 degrees and cos 72 degrees using Parts a-c. (Do not use a calculator!)

I got part (a), you can easily find and use the missing angles to prove BC = BD = AD. But I need a bit of help with the other parts.

Thanks!

Thank you so much!

(i think I figured out the third... still working on it though!)

**championmathgirl**- Replies: 3

Why does cot^2(x) + 1 = csc^2(x) for any x such that x is not an integer multiple of 180 degrees, where cot=cotangent and csc=cosecant?

Cotangent=(cos(x))/(sin(x)) and Cosecant=1/(sin(x))

My understanding is (cos^2(x))/(sin^2(x))+1=1/(sin^2(x)), multiplying the equation by sin^2x to get;

cos^2(x)+sin^2(x)=sin^2(x), but that doesn't get me anywhere with why that's true for all x that's not a multiple of 180. A little help please?

Thanks

Those are great proofs. But is there a way using coordinate geometry? For example putting triangle ABC in the coordinate plane so that A is the origin and one of the equal sides aligns with the x-axis.

I tried to go on from there, but I'm not sure what to do.

bob bundy wrote:

Imagine rotating triangle TAC through 90 degrees, anticlockwise, around point A.

T rotates to B and C rotates to V, so one triangle becomes the other.

So TC must also rotate 90 to become BV, hence they are perpendicular.

Bob

How do we know that you have to rotate it 90 degrees?

I got 182pi, is that right?

bob bundy wrote:

To get the ratio of volumes you cube the length SF.

I'm also working on this problem.

Where did the "S" come from?

**championmathgirl**- Replies: 7

Let the incircle of triangle

be tangent to sides , , and at , , and , respectively. Prove that triangle is acute., I'm new to this forum, I don't know whether the diagram came out.

I got a hint saying to compute the angles of triangle

in terms of the angles of triangle , but after doing so, got me nowhere.If I could get some help, that would be great!

@bob bundy, I was wondering how you got angle DCB = 180 - (x + y) => BCQ = x + y.

Pages: **1**