Properties of the Circumcentre of a Title Triangle
When you are still in high school, you encounter the problem of the circumcenter of a triangle, are you worried?
Then I will give you a trick today, hhhhhhh
16340129 School of Data Science and Computer
First, the triangle circumcenter problem?
1. This is not much to say, there are problems with pictures, and such problems are generally related to vectors
It involves the intersection of the perpendicular bisectors of the three sides of the triangle at the outer center O, and here we introduce the key to solving the problem.
Two, official
1. Form
If the circumcenter O of \(\Delta\)ABC, then \(\Delta\)ABC has the following properties:
\ (\ vec {AO} \) \ (\ cdot \) \ (\ vec {AB} \) = \ (\ frac {1} {2} \) \ (\ vec {AB ^ {2}} \)
\ (\ vec {AO} \) \ (\ cdot \) \ (\ vec {AC} \) = \ (\ frac {1} {2} \) \ (\ vec {AC ^ {2}} \)
Proof: According to the definition of vector inner product, when \(\vec{AO}\) is projected onto \(\vec{AB}\), it is the midpoint of line segment AB, which is proved.
Third, the solution?
1. Solution:
\ (\ vec {AO} \) \ (\ cdot \) \ (\ vec {AB} \) = \ (\ frac {1} {2} \) \ (\ vec {AB ^ {2}} \) = 18 -------- ①
\ (\ vec {AO} \) \ (\ cdot \) \ (\ vec {AC} \) = \ (\ frac {1} {2} \) \ (\ vec {AC ^ {2}} \) = 50 -------- ②
From the title: \(\vec{AO}\)=x\(\cdot\)\(\vec{AB}\)+y\(\cdot\)\(\vec{AC}\)
Bring in ①② and get: \(\vec{AO}\)\(\cdot\)\(\vec{AB}\)=x\(\vec{AB^{2}}\)+y\(\ vec{AB}\)\(\cdot\)\(\vec{AC}\)=18=36x+y\(\cdot\)\(\vec{AB}\)\(\cdot\)\( \vec{AB}\)
The same can be obtained: \(\vec{AO}\)\(\cdot\)\(\vec{AC}\)=100y+x\(\vec{AB}\)\(\cdot\)\( \vec{AB}\)
Set \(\vec{AB}\)\(\cdot\)\(\vec{AC}\)=m, solve the following inequality: 36x+ym=18
100y+xm=50
2x+10y=5
Solution: \(m_1\)=20,\(m_2\)=36
The following is a classification discussion: ①When \(m_1\)=20, \(\vec{AB}\)\(\cdot\)\(\vec{AC}\)=20, \(\cos\)∠BAC =\(\frac{1}{2}\)
②当\(m_2\)=36，\(\vec{AB}\)\(\cdot\)\(\vec{AC}\)=36，\(\cos\)∠BAC=\(\frac{3}{5}\)
To sum up: \(\cos\)∠BAC=\(\frac{1}{2}\) or \(\cos\)∠BAC=\(\frac{3}{5}\)
[footnote]
[1] The title is adapted from Zhuhai No. 1 Middle School and No. 6 School Triple Link