Geometry-P3

Problem↑

Solution (Several ways):

Make CE⊥AP, Connect BE.

Since m∠CEP=90° and m∠EPC=60

m∠ECP=30°

According to the 30-60-90 triangle theorem

2EP=PC

since 2BP=PC

EP=BP

m∠EBP=m∠BEP (EP's oppsite angle is congruent to BP's)

since m∠APC=60

m∠APB=120

m∠EBP=m∠BEP=(180-120)/2=30

Since m∠EBP=m∠ECP

BE=EC

m∠EBP=45-30=15

m∠BAP=60-45=15 (exterior angles theorem)

Since m∠EBP=m∠BAP

BE=AE=EC

Since AE=EC and ∠AEC=90

∠EAC=∠ACE=45

Since m∠ACE+m∠ECP=m∠C (angles addition postulate)

m∠C=45+30=75°

Geometry-P2

Problem:
An interior angle is missing in a convex polygon that has n sides, the sum of the other interior angles is 2570
find the value of n (how many sides this polygon has).

Solution:
The polygon interior angles theorem: 
The sum of the measures of the interior angles: (n-2)*180
Let x=the measure of the missing interior angle
(n-2)*180=2570+x
x=180n-2930
Since this is a convex polygon
0<x<180
therefore:
0<180n-2930<180
2930<180n<3110
16.2777<n<17.27777
Since n has to be a natural number
n has to be equal to 17
Therefore n=17

Interesting stuffs

1.   1/a+1/b=(a+b)/ab
Explanation:
1/a*b/b=b/ab
1/b*a/a=a/ab
a/ab+b/ab=(a+b)/ab

2.  
When x-y≥0
and y-x≥0
x=y
Explanation:
x≥y
y≥x
x=y (Only choice)          

Geometry-The Euclid Theorem Part1

The Euclid Theorem, also known as  射影定理, can be used when there's an triangle like this (Not really): 

                                                                                          

As can be seen, m∠ABC=90° and BD⊥AC

The following statements are true based on the given information.

1. (BD)^2=AD*DC，2.(AB)^2=AD*AC ，3.(BC)^2=CD*CA , 4.AB*BC=AC*BD

Proof for the the first statement: 

Since m∠BDC=90

m∠DBC+m∠C=90 (corollary to the triangle sum theorem)

and SInce m∠ABC=90

m∠A+m∠C=90   (corollary to the triangle sum theorem)

m∠DBC=m∠A

for the same reason

m∠A+m∠ABD=90

m∠A+m∠C=90

m∠C=m∠ABD

Since ∠C≌∠ABD and m∠BDA=m∠BDC=90

By the Angle-Angle Similarity Postulate

△ABD∽△BCD

Therefore AD/BD=BD/CD

(BD)^2=AD*DC

Geometry-P1

Since CF bisects∠ACB, BE bisect∠ABC
m∠CBO=m∠ABC/2
m∠BCO=m∠ACB/2
Using the Triangle Sum Theorem
m∠CBO+m∠BCO+m∠BOC=180
m∠ABC/2+m∠ACB/2+m∠BOC=180
(m∠ABC+m∠ACB)/2+m∠BOC=180
Since 180-m∠A=(m∠ABC+m∠ACB)
(m∠ABC+m∠ACB)=116
116/2+m∠BOC=180
58+m∠BOC=180
m∠BOC=122              

Inequality-P1

Problem:


Given that x>1
Find the lowest value of x+4/(x-1)


Solution:


x>1
x-1>0
let a=(x-1) b=4/(x-1)
Since (√a-√b)^2>0 and a≠b
a-2√ab+b>0
a+b>2√ab
By plugging in a and b
(x-1)+4/(x-1)>2√4(x-1)/(x-1)=4
(x-1)+4/(x-1)>4
Since (x-1)+4/(x-1)+1=x+4/(x-1)
x+4/(x-1)>5