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#1 2015-12-21 06:16:40

mathster
Member
Registered: 2015-10-09
Posts: 15

Geometry problems.

In the diagram below, we have $\angle ABC = \angle ACB = \angle DEC=\angle CDE$, $BC = 8$, and $DB = 2$. Find $AB$.

Angle bisectors $\overline{TX}$ and $\overline{UY}$ of $\triangle TUV$ meet at point $I$. Find all possible values of $\angle V$ if $\angle  TIU = 109^\circ$. As your answer, enter the number of degrees in $\angle V$. If you find more than one possibility, list the possible values in increasing order, separated by commas.

The interior angle measures of a pentagon form an arithmetic progression. The difference between the largest and smallest angle measures is $44^\circ$. Find the measure of the smallest angle, in degrees.

Let $ABCD$ be a convex quadrilateral, and let $P$, $Q$, $R$, $S$, $T$, $U$, $V$, and $W$ be the trisection points of the sides of $ABCD$, as shown.
http://latex.artofproblemsolving.com/8/3/d/83df21d419d8743fb5b781faac136a6c8676d4ef.png
If the area of quadrilateral $ABCD$ is 180, then find the area of hexagon $AQRCUV$.

In the triangle shown, $n$ is a positive integer, and $\angle A > \angle B > \angle C$. How many possible values of $n$ are there?

d4c0ef5eaa79738f8864060f1490f3a9f85b0939.png

Help very greatly appreciated!

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#2 2015-12-21 07:29:24

Bob
Administrator
Registered: 2010-06-20
Posts: 10,621

Re: Geometry problems.

hi mathster,

Firstly, why are you embedding $ and \ in your post.  These do nothing on my screen except confuse the real text.  I recommend you leave them out in future.  Hopefully, I've got the true sense of these questions.

Q1.  Cannot help with this as I see no diagram.  If ABC is a triangle, where are D and E?

Q2. I've called angle TUV = 2x and angle UTV = 2y.

Using the angle sum of a triangle you can work out x + y, then IYV and IXV in terms of x and y.  Then using the angle sum of a quadrilateral you can compute the angle at V.  As far as I can tell, there is only one answer.

Q3.  Call the angles a, a+d, a+2d,a+3d and a+4d.  The sum is 540 and the given information will enable you to work out d.  So all you have to do is work out 'a'.

Q4.  Those trisections mean that triangles ACVD, WTD and VUD are similar.  Also ACB, PSB and QRB.

If you call AC = 3k and the height of ACB = 3H and the height of ACD = 3G, then you can express the area of the quadrilateral in terms of K, H and G.  Thus you can state the value of KH + KG.

The area of triangles QRB and UVD can also be expressed in similar terms, and so, by subtraction, you can get the area of the hexagon.

Q5.  In any triangle the longest side is opposite the largest angle, middle sized side opposite the middle sized angle, and smallest side opposite the smallest angle.  Thus you can deduce that

3n+4 > 3n+1 > 4n-9

The first pair gives no help, but first with third and also second with third lead to an upper bound for n.

A lower bound comes from knowing that no side can have a negative length.

Hope that helps,

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2015-12-21 08:52:29

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Geometry problems.

Firstly, why are you embedding $ and \ in your post.

The $ character is used to denote latex on other sites.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#4 2015-12-21 21:44:44

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Geometry problems.

Just passing through,
For 5., bob has it, and the range is

Last edited by Relentless (2015-12-21 21:46:28)

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