polygons and triangles

**Raiden_11** · July 4th 2007, 10:52 PM

**red_dog** · July 4th 2007, 11:41 PM

Question 1:
$AC=\sqrt{AB^2+AC^2}=\sqrt{2}$
$AD=\sqrt{AC^2+CD^2}=\sqrt{2+1}=\sqrt{3}$
$AE=\sqrt{AD^2+DE^2}=\sqrt{3+1}=2$
$AF=\sqrt{AE^2+EF^2}=\sqrt{4+1}=\sqrt{5}$

**red_dog** · July 4th 2007, 11:45 PM

Question 2:
$\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}\Ri ghtarrow 2(x+y+z)=180^{\circ}\Rightarrow x+y+z=90^{\circ}$
$\widehat{ADB}=180^{\circ}-2z=180^{\circ}-2(90^{\circ}-(x+y))=$
$=180^{\circ}-180^{\circ}+2(x+y)=2\widehat{ACB}$

**red_dog** · July 4th 2007, 11:50 PM

Question 3:
If $\displaystyle \frac{AC}{AB}=k$ then $\displaystyle \overrightarrow{OM}=\frac{1}{1+k}\cdot\overrightar row{OC}+\frac{k}{1+k}\cdot\overrightarrow{OB}$
In this case $\displaystyle k=\frac{7}{3}$

**red_dog** · July 4th 2007, 11:57 PM

Question 4:
$\widehat{FBG}$ is exterior to triangle $BDG$ . So $\widehat{FBG}>\widehat{BDG}$ (1).
$\widehat{BDG}$ is exterior to triangle $ADE$ . So $\widehat{BDG}>\widehat{AED}$ (2).
From (1) and (2) via tranzitivity, yields $\widehat{FBC}>\widehat{AED}$

**earboth** · July 5th 2007, 05:10 AM

Hello,

to question 2:

I've modified your drawing (see attachment):

The angle in red is an exterior angle of the isosceles triangle ACD;
the angle in blue is an exterior angle of the isosceles triangle ABD.

The angle at the centre is 2x + 2y = 2(x+y) that means it is twice as large as the angle at C.

**Soroban** · July 5th 2007, 06:21 AM

Hello, Raiden_11!

Basically the same as red_dog . . .

Click image for larger version.

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In $\Delta ABC\!:\;2x + 2y + 2z \;=\;180^o\quad\Rightarrow\quad x + y + z \;=\;90^o$

. . $\angle BCA \;=\;x + y \;=\;90^o - z$

In $\Delta ADB\!:\;\angle ADB + 2z \;=\;180^o$

Therefore: . $\angle ADB \;=\;180 - 2z \;=\;2(90^o-z) \;=\;2\cdot\angle BCA$

**Raiden_11** · July 5th 2007, 10:02 AM

2. Observe the relationship between the elements of Diagonal 1 and Diagonal 2 in Pascal’s Triangle, shown on the next page. The sum of the first two elements in Diagonal 1 is equal to the second element in Diagonal 2. Similarly, the sum of the first 5 elements of Diagonal 1 is equal to the 5th element of Diagonal 2. Suppose we wish to use Pascal’s Triangle to find the sum of the first n natural numbers. Use combinatorial notation to express the sum of the n elements of Diagonal 1 in terms of the corresponding value in Diagonal 2.

3. Using the method of mathematical induction, prove that the following statements are false.

**red_dog** · July 5th 2007, 10:22 AM

2.
$C_1^1+C_2^1=C_3^2$
$C_1^1+C_2^1+C_3^1+C_4^1+C_5^1=C_6^2$
$C_1^1+C_2^1+\ldots C_n^1=C_{n+1}^2$

3.
For the first identity:
For $n=1\Rightarrow 2=2$
For $n=2\Rightarrow 6=6$
For $n=3\Rightarrow 12=14$ , false. So the identity is not true for all $n\in\mathbf{N^*}$ .

For the second identity:
For $n=1\Rightarrow 2=4$ , false. So the identity is not true for all $n\in\mathbf{N^*}$ .

**Raiden_11** · July 5th 2007, 11:27 AM

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