Given points
Let
A ( 6 , 8 ) , B ( 10 cos α , − 10 sin α ) , C ( − 10 sin α , 10 cos α ) . A(6,8), \quad B(10\cos\alpha,-10\sin\alpha), \quad C(-10\sin\alpha,10\cos\alpha). A ( 6 , 8 ) , B ( 10 cos α , − 10 sin α ) , C ( − 10 sin α , 10 cos α ) .
We need to find
5 a − 3 h + 6 k + 100 sin 2 α , 5a-3h+6k+100\sin 2\alpha, 5 a − 3 h + 6 k + 100 sin 2 α , L ( a , 9 ) L(a,9) L ( a , 9 ) G ( h , k ) G(h,k) G ( h , k ) A B C ABC A B C
Find the centroid G ( h , k ) G(h,k) G ( h , k )
The centroid is the average of the coordinates of the vertices:
h = 6 + 10 cos α − 10 sin α 3 , h=\frac{6+10\cos\alpha-10\sin\alpha}{3}, h = 3 6 + 10 c o s α − 10 s i n α , k = 8 − 10 sin α + 10 cos α 3 . k=\frac{8-10\sin\alpha+10\cos\alpha}{3}. k = 3 8 − 10 s i n α + 10 c o s α .
So,
G ( 6 + 10 cos α − 10 sin α 3 , 8 − 10 sin α + 10 cos α 3 ) . G\left(\frac{6+10\cos\alpha-10\sin\alpha}{3},\frac{8-10\sin\alpha+10\cos\alpha}{3}\right). G ( 3 6 + 10 c o s α − 10 s i n α , 3 8 − 10 s i n α + 10 c o s α ) .
Use the fact that L ( a , 9 ) L(a,9) L ( a , 9 )
If L ( a , 9 ) L(a,9) L ( a , 9 )
A L ⊥ B C AL \perp BC A L ⊥ B C B L ⊥ A C BL \perp AC B L ⊥ A C
We will use these conditions.
Slope of B C BC B C A L ⊥ B C AL \perp BC A L ⊥ B C
First compute direction vector of B C BC B C B C → = ( − 10 sin α − 10 cos α , 10 cos α + 10 sin α ) . \overrightarrow{BC}=\big(-10\sin\alpha-10\cos\alpha,\;10\cos\alpha+10\sin\alpha\big). B C = ( − 10 sin α − 10 cos α , 10 cos α + 10 sin α ) .
This is proportional to
( − 1 , 1 ) , (-1,1), ( − 1 , 1 ) , B C → = 10 ( sin α + cos α ) ( − 1 , 1 ) . \overrightarrow{BC}=10(\sin\alpha+\cos\alpha)(-1,1). B C = 10 ( sin α + cos α ) ( − 1 , 1 ) .
Hence slope of B C BC B C m B C = − 1. m_{BC}=-1. m B C = − 1.
Therefore altitude from A A A 1 1 1
Since A = ( 6 , 8 ) A=(6,8) A = ( 6 , 8 ) L = ( a , 9 ) L=(a,9) L = ( a , 9 ) slope of A L = 9 − 8 a − 6 = 1 a − 6 . \text{slope of }AL=\frac{9-8}{a-6}=\frac{1}{a-6}. slope of A L = a − 6 9 − 8 = a − 6 1 . 1 1 1 1 a − 6 = 1 ⟹ a − 6 = 1 ⟹ a = 7. \frac{1}{a-6}=1 \implies a-6=1 \implies a=7. a − 6 1 = 1 ⟹ a − 6 = 1 ⟹ a = 7.
Now compute the required expression using centroid
We need
5 a − 3 h + 6 k + 100 sin 2 α . 5a-3h+6k+100\sin 2\alpha. 5 a − 3 h + 6 k + 100 sin 2 α .
Since a = 7 a=7 a = 7 5 a = 35. 5a=35. 5 a = 35.
Now,
− 3 h = − ( 6 + 10 cos α − 10 sin α ) , -3h=-(6+10\cos\alpha-10\sin\alpha), − 3 h = − ( 6 + 10 cos α − 10 sin α ) , 6 k = 2 ( 8 − 10 sin α + 10 cos α ) = 16 − 20 sin α + 20 cos α . 6k=2(8-10\sin\alpha+10\cos\alpha)=16-20\sin\alpha+20\cos\alpha. 6 k = 2 ( 8 − 10 sin α + 10 cos α ) = 16 − 20 sin α + 20 cos α .
Add these:
5 a − 3 h + 6 k = 35 − ( 6 + 10 cos α − 10 sin α ) + 16 − 20 sin α + 20 cos α . 5a-3h+6k = 35-(6+10\cos\alpha-10\sin\alpha)+16-20\sin\alpha+20\cos\alpha. 5 a − 3 h + 6 k = 35 − ( 6 + 10 cos α − 10 sin α ) + 16 − 20 sin α + 20 cos α .
Simplify:
= 35 − 6 − 10 cos α + 10 sin α + 16 − 20 sin α + 20 cos α , =35-6-10\cos\alpha+10\sin\alpha+16-20\sin\alpha+20\cos\alpha, = 35 − 6 − 10 cos α + 10 sin α + 16 − 20 sin α + 20 cos α , = 45 + 10 cos α − 10 sin α . =45+10\cos\alpha-10\sin\alpha. = 45 + 10 cos α − 10 sin α .
So the full expression is
45 + 10 cos α − 10 sin α + 100 sin 2 α . 45+10\cos\alpha-10\sin\alpha+100\sin 2\alpha. 45 + 10 cos α − 10 sin α + 100 sin 2 α .
This still contains α \alpha α
Use B L ⊥ A C BL \perp AC B L ⊥ A C
Direction vector of A C AC A C A C → = ( − 10 sin α − 6 , 10 cos α − 8 ) . \overrightarrow{AC}=(-10\sin\alpha-6,\;10\cos\alpha-8). A C = ( − 10 sin α − 6 , 10 cos α − 8 ) .
Direction vector of B L BL B L B L → = ( 7 − 10 cos α , 9 + 10 sin α ) . \overrightarrow{BL}=(7-10\cos\alpha,\;9+10\sin\alpha). B L = ( 7 − 10 cos α , 9 + 10 sin α ) .
Since B L ⊥ A C BL \perp AC B L ⊥ A C B L → ⋅ A C → = 0. \overrightarrow{BL}\cdot \overrightarrow{AC}=0. B L ⋅ A C = 0.
So,
( 7 − 10 cos α ) ( − 10 sin α − 6 ) + ( 9 + 10 sin α ) ( 10 cos α − 8 ) = 0. (7-10\cos\alpha)(-10\sin\alpha-6)+(9+10\sin\alpha)(10\cos\alpha-8)=0. ( 7 − 10 cos α ) ( − 10 sin α − 6 ) + ( 9 + 10 sin α ) ( 10 cos α − 8 ) = 0.
Expand:
− 70 sin α − 42 + 100 sin α cos α + 60 cos α + 90 cos α − 72 + 100 sin α cos α − 80 sin α = 0. -70\sin\alpha-42+100\sin\alpha\cos\alpha+60\cos\alpha+90\cos\alpha-72+100\sin\alpha\cos\alpha-80\sin\alpha=0. − 70 sin α − 42 + 100 sin α cos α + 60 cos α + 90 cos α − 72 + 100 sin α cos α − 80 sin α = 0.
Combine terms:
200 sin α cos α + 150 cos α − 150 sin α − 114 = 0. 200\sin\alpha\cos\alpha +150\cos\alpha -150\sin\alpha -114=0. 200 sin α cos α + 150 cos α − 150 sin α − 114 = 0.
Using
2 sin α cos α = sin 2 α , 2\sin\alpha\cos\alpha=\sin 2\alpha, 2 sin α cos α = sin 2 α , 100 sin 2 α + 150 cos α − 150 sin α − 114 = 0. 100\sin 2\alpha +150\cos\alpha -150\sin\alpha -114=0. 100 sin 2 α + 150 cos α − 150 sin α − 114 = 0.
Thus,
100 sin 2 α = 114 − 150 cos α + 150 sin α . 100\sin 2\alpha =114-150\cos\alpha+150\sin\alpha. 100 sin 2 α = 114 − 150 cos α + 150 sin α .
Substitute into the expression
Required expression:
45 + 10 cos α − 10 sin α + 100 sin 2 α . 45+10\cos\alpha-10\sin\alpha+100\sin 2\alpha. 45 + 10 cos α − 10 sin α + 100 sin 2 α .
Substitute the value of 100 sin 2 α 100\sin 2\alpha 100 sin 2 α = 45 + 10 cos α − 10 sin α + 114 − 150 cos α + 150 sin α . =45+10\cos\alpha-10\sin\alpha+114-150\cos\alpha+150\sin\alpha. = 45 + 10 cos α − 10 sin α + 114 − 150 cos α + 150 sin α .
Simplify:
= 159 − 140 cos α + 140 sin α . =159-140\cos\alpha+140\sin\alpha. = 159 − 140 cos α + 140 sin α .
This seems not constant yet, so let us instead use the orthocenter relation more directly via the standard property:
For a triangle with circumcenter at origin, orthocenter position vector satisfies
O H ⃗ = O A ⃗ + O B ⃗ + O C ⃗ . \vec{OH}=\vec{OA}+\vec{OB}+\vec{OC}. O H = O A + O B + O C .
Let us check whether the three points lie on the circle centered at origin.
For A ( 6 , 8 ) A(6,8) A ( 6 , 8 ) O A = 6 2 + 8 2 = 10. OA=\sqrt{6^2+8^2}=10. O A = 6 2 + 8 2 = 10. B B B O B = 10. OB=10. O B = 10. C C C O C = 10. OC=10. O C = 10.
Hence all three points lie on the circle
x 2 + y 2 = 100 , x^2+y^2=100, x 2 + y 2 = 100 , L = H = A + B + C . L=H=A+B+C. L = H = A + B + C .
So,
H = ( 6 + 10 cos α − 10 sin α , 8 − 10 sin α + 10 cos α ) . H=(6+10\cos\alpha-10\sin\alpha,\;8-10\sin\alpha+10\cos\alpha). H = ( 6 + 10 cos α − 10 sin α , 8 − 10 sin α + 10 cos α ) .
But L = ( a , 9 ) L=(a,9) L = ( a , 9 ) y y y 8 − 10 sin α + 10 cos α = 9 , 8-10\sin\alpha+10\cos\alpha=9, 8 − 10 sin α + 10 cos α = 9 , 10 cos α − 10 sin α = 1 , 10\cos\alpha-10\sin\alpha=1, 10 cos α − 10 sin α = 1 , 10 ( cos α − sin α ) = 1. 10(\cos\alpha-\sin\alpha)=1. 10 ( cos α − sin α ) = 1.
Also,
a = 6 + 10 cos α − 10 sin α = 6 + 1 = 7. a=6+10\cos\alpha-10\sin\alpha=6+1=7. a = 6 + 10 cos α − 10 sin α = 6 + 1 = 7.
This is consistent.
Now centroid is
G = ( 6 + 10 cos α − 10 sin α 3 , 8 − 10 sin α + 10 cos α 3 ) = ( 7 3 , 3 ) . G=\left(\frac{6+10\cos\alpha-10\sin\alpha}{3},\frac{8-10\sin\alpha+10\cos\alpha}{3}\right)=\left(\frac73,3\right). G = ( 3 6 + 10 c o s α − 10 s i n α , 3 8 − 10 s i n α + 10 c o s α ) = ( 3 7 , 3 ) .
Thus,
h = 7 3 , k = 3 , a = 7. h=\frac73, \quad k=3, \quad a=7. h = 3 7 , k = 3 , a = 7.
Now evaluate:
5 a − 3 h + 6 k + 100 sin 2 α . 5a-3h+6k+100\sin 2\alpha. 5 a − 3 h + 6 k + 100 sin 2 α .
First,
5 a − 3 h + 6 k = 5 ( 7 ) − 3 ⋅ 7 3 + 6 ( 3 ) = 35 − 7 + 18 = 46. 5a-3h+6k=5(7)-3\cdot \frac73 +6(3)=35-7+18=46. 5 a − 3 h + 6 k = 5 ( 7 ) − 3 ⋅ 3 7 + 6 ( 3 ) = 35 − 7 + 18 = 46.
Now find 100 sin 2 α 100\sin 2\alpha 100 sin 2 α
We have
cos α − sin α = 1 10 . \cos\alpha-\sin\alpha=\frac{1}{10}. cos α − sin α = 10 1 .
Use
( cos α − sin α ) 2 = 1 − 2 sin α cos α = 1 − sin 2 α . (\cos\alpha-\sin\alpha)^2=1-2\sin\alpha\cos\alpha=1-\sin 2\alpha. ( cos α − sin α ) 2 = 1 − 2 sin α cos α = 1 − sin 2 α .
So,
( 1 10 ) 2 = 1 − sin 2 α , \left(\frac{1}{10}\right)^2 = 1-\sin 2\alpha, ( 10 1 ) 2 = 1 − sin 2 α , 1 100 = 1 − sin 2 α , \frac{1}{100}=1-\sin 2\alpha, 100 1 = 1 − sin 2 α , sin 2 α = 99 100 . \sin 2\alpha=\frac{99}{100}. sin 2 α = 100 99 .
Therefore,
100 sin 2 α = 99. 100\sin 2\alpha=99. 100 sin 2 α = 99.
Finally,
5 a − 3 h + 6 k + 100 sin 2 α = 46 + 99 = 145. 5a-3h+6k+100\sin 2\alpha=46+99=145. 5 a − 3 h + 6 k + 100 sin 2 α = 46 + 99 = 145.
Final answer
145 \boxed{145} 145