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Given the points $ A(1, 0) $, $ B(5, 0) $, and $ C(0, 5) $, a circle is drawn with center at point $ B $ and radius 2. Let $ P $ be a moving point on this circle. I need to find the minimum value of $ PC + \frac{1}{2}PA $, where $ P $ is the point on the circle.

Any insights or alternative approaches would be greatly appreciated! enter image description here

Here's a solution. enter image description here To find the minimum value of $ PC + \frac{1}{2}PA $ geometrically, we can use the following approach:

  1. Identify Points and Segments:

    • Given points: $A(1, 0)$, $B(5, 0)$, and $C(0, 5)$.
    • Circle centered at $B$ with radius $2$.
    • Let $P$ be a point on the circle.
    • Select point $ F $ on segment $ AB $ such that $ BF = 1 $.

  2. Triangles and Similarity:

    • Since $ PB = 2 $ and $ BF = 1 $: $$ \frac{BF}{PB} = \frac{1}{2} $$
    • Triangles $ \triangle PBF $ and $ \triangle ABP $ are similar because: $$ \frac{PB}{AB} = \frac{2}{4} = \frac{1}{2} = \frac{BF}{PB} $$ Therefore, by AA similarity ($\angle PBF = \angle ABP$), we have: $$ \triangle PBF \sim \triangle ABP $$

  3. Proportional Segments:

    • Since the triangles are similar: $$ \frac{PF}{PA} = \frac{BF}{PB} = \frac{1}{2} \implies PF = \frac{1}{2}PA $$

  4. Inequality and Minimum Value:

    • From the above proportional relationship: $$ PC + \frac{1}{2}PA = PC + PF \geq CF $$
    • When $ C $, $ P $, and $ F $ are collinear, $ PC + \frac{1}{2}PA $ is minimized, and this minimum value is the length of segment $ CF $.

  5. Calculate ( CF ):

    • Since $ O $ is the origin, $ OC = 5 $ and $ OF = OB - 1 = 5 - 1 = 4 $: $$ CF = \sqrt{OC^2 + OF^2} = \sqrt{5^2 + 4^2} = \sqrt{41} $$

Thus, the minimum value of $ PC + \frac{1}{2}PA $ is: $$ \sqrt{41} $$

The limitation of this method is that if the coordinates of key point A change, and we still need to find the minimum value of $PC + \frac{1}{2}PA$, this method will no longer be applicable.

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    $\begingroup$ Parametrizing the point $P$ as $P = (5 + 2 \cos t , 2 \sin t) $, and expressing the required distance in terms of $t$, then directly differentiating that, and finding the critical points, I found that the minimum is $\sqrt{41}= 6.40312423$, and it occurs at $ t = 2.64664 \text{ radian } = 151.6413^\circ $ $\endgroup$
    – Quadrics
    Commented May 24 at 17:53

4 Answers 4

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enter image description here

To find the minimum value of $ PC + \frac{1}{2}PA $ geometrically, we can use the following approach:

  1. Identify Points and Segments:

    • Given points: $A(1, 0)$, $B(5, 0)$, and $C(0, 5)$.
    • Circle centered at $B$ with radius $2$.
    • Let $P$ be a point on the circle.
    • Select point $ F $ on segment $ AB $ such that $ BF = 1 $.

  2. Triangles and Similarity:

    • Since $ PB = 2 $ and $ BF = 1 $: $$ \frac{BF}{PB} = \frac{1}{2} $$
    • Triangles $ \triangle PBF $ and $ \triangle ABP $ are similar because: $$ \frac{PB}{AB} = \frac{2}{4} = \frac{1}{2} = \frac{BF}{PB} $$ Therefore, by AA similarity ($\angle PBF = \angle ABP$), we have: $$ \triangle PBF \sim \triangle ABP $$

  3. Proportional Segments:

    • Since the triangles are similar: $$ \frac{PF}{PA} = \frac{BF}{PB} = \frac{1}{2} \implies PF = \frac{1}{2}PA $$

  4. Inequality and Minimum Value:

    • From the above proportional relationship: $$ PC + \frac{1}{2}PA = PC + PF \geq CF $$
    • When $ C $, $ P $, and $ F $ are collinear, $ PC + \frac{1}{2}PA $ is minimized, and this minimum value is the length of segment $ CF $.

  5. Calculate ( CF ):

    • Since $ O $ is the origin, $ OC = 5 $ and $ OF = OB - 1 = 5 - 1 = 4 $: $$ CF = \sqrt{OC^2 + OF^2} = \sqrt{5^2 + 4^2} = \sqrt{41} $$

Thus, the minimum value of $ PC + \frac{1}{2}PA $ is: $$ \sqrt{41} $$

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  • $\begingroup$ I just found here a recent answer to a question which is (almost) exactly the same as the present one and the solution by Hwang is in the very same spirit as yours. $\endgroup$
    – Jean Marie
    Commented May 24 at 20:50
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    $\begingroup$ Also of interest is the fact that this issue is a generalization of Alhazen problem which can be solved... using apollonian circles. $\endgroup$
    – Jean Marie
    Commented May 24 at 21:30
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As you are asking for insights, here is a method able to give you a graphical approximate answer.

Its principle relies on the fact that the set of points $M$ such that

$$MC + \frac{1}{2}MA=c$$

(where $c$ is a constant) is a closed curve that can be called a "generalized ellipse". We have featured some of these curves on the following figure, here ranging from $c=3$, the smallest, to $c=6.4$ the largest (with $0.2$ steps). This largest value corresponds to a case of approximate tangency with the circle in a point $P$ which is the desired point. This "numerically optimal" value

$$c \ \approx \ 6.4 \ \text{is indeed very close to } \ \sqrt{41} \ (\approx 6.403...)$$

It is probably possible to use this method to derive the exact value in a rigorous way by expressing the proportionality of gradients $G_1$ and $G_2$ in the optimal point of contact :

$$\frac{\vec{PA}}{\|PA\|}+\tfrac12 \frac{\vec{PC}}{\|PC\|}= \lambda \vec{PB}$$

(where $\lambda$ is a Lagrange multiplier) but till now, I haven't succeeded.

Moreover, it is difficult to compete with the nice solution of @Oth S...

enter image description here

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  • $\begingroup$ @Intelligenci pauca Thanks a lot. I have made the changes. $\endgroup$
    – Jean Marie
    Commented May 24 at 19:46
  • $\begingroup$ @Intelligenci pauca Please, see the comments I have made to Oth S $\endgroup$
    – Jean Marie
    Commented May 24 at 21:24
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Here is a derivation using SymPy, a CAS in Python:

# define objects

>>> from sympy.abc import *
>>> from sympy import *
>>> P=Circle((5,0),2).arbitrary_point(t)
>>> C=Point(0,5)
>>> A=Point(1,0)

# define objective

>>> objective=Segment(P,C).length + Segment(P,A).length/2
sqrt((2*sin(t) - 5)**2 + (2*cos(t) + 5)**2) + sqrt((2*cos(t) + 4)**2 + 4*sin(t)**2)/2

# we desire the minimum wrt t (which controls where we are on the circle)

>>> objective.diff(t)
(2*(2*sin(t) - 5)*cos(t) - 2*(2*cos(t) + 5)*sin(t))/sqrt((2*sin(t) - 5)**2 + (2*cos(t) + 5)**2) + (-2*(2*cos(t) + 4)*sin(t) + 4*sin(t)*cos(t))/(2*sqrt((2*cos(t) + 4)**2 + 4*sin(t)**2))

# rewrite sin and cos in terms of tan and let tan(t/2) be x, keep only
# the numerator

>>> _.rewrite(tan).subs(tan(t/2),x).as_numer_denom()[0]
(x**2 + 1)**2*(8*x*(1 - x**2) - 8*x*(x**2 + 3))*sqrt((4*x/(x**2 + 1) - 5)**2 + (2*(1 - x**2)/(x**2 + 1) + 5)**2) + (x**2 + 1)**2*(-8*x*(3*x**2 + 7) + 4*(1 - x**2)*(-5*x**2 + 4*x - 5))*sqrt(16*x**2/(x**2 + 1)**2 + (2*(1 - x**2)/(x**2 + 1) + 4)**2)

# remove radicals

>>> unrad(_)
(25*x**22 - 100*x**21 + 339*x**20 - 1440*x**19 + 1991*x**18 - 7020*x**17 + 6949*x**16 - 16320*x**15 + 16346*x**14 - 17640*x**13 + 27454*x**12 + 33614*x**10 + 24360*x**9 + 29866*x**8 + 31680*x**7 + 18749*x**6 + 19980*x**5 + 7871*x**4 + 6560*x**3 + 1979*x**2 + 900*x + 225, [])

# factor

>>> factor(_[0])
(x**2 + 1)**8*(5*x**2 - 16*x - 15)*(5*x**4 - 4*x**3 + 30*x**2 - 44*x - 15)

# get solutions for x

>>> [i for i in solve(_) if i.is_real]
[8/5 - sqrt(139)/5, 8/5 + sqrt(139)/5, 1/5 + sqrt(-96/25 - 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3))/2 + sqrt(-192/25 - 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3) + 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 1616/(125*sqrt(-96/25 - 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3))))/2, -sqrt(-192/25 - 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3) + 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 1616/(125*sqrt(-96/25 - 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3))))/2 + 1/5 + sqrt(-96/25 - 88/(75*(196/25 + 4*sqrt(4877997)/1125)**(1/3)) + 2*(196/25 + 4*sqrt(4877997)/1125)**(1/3))/2]

>>> ans = flatten([solve(tan(t/2)-i) for i in _])

# check value of objective at those values of t

>>> [objective.subs(t,i).n() for i in ans]
[11.3350373276846, 6.40312423743285, 7.11408182438962, 11.9279919892822]

# value of objective at ans[1]

>>> m=objective.subs(t,ans[1]);m
sqrt(4*sin(2*atan(8/5 + sqrt(139)/5))**2 + (2*cos(2*atan(8/5 + sqrt(139)/5)) + 4)**2)/2 + sqrt((2*cos(2*atan(8/5 + sqrt(139)/5)) + 5)**2 + (-5 + 2*sin(2*atan(8/5 + sqrt(139)/5)))**2)

# looks worse that it is - this is equivalent to sqrt(41)

>>> nsimplify(m)
sqrt(41)
>>> m.equals(_)
True

So the minimum is $\sqrt{41}$ at a value of $\tan(t/2) =(8/5 + \sqrt{139}/5)$. This agrees with the graphical minimum of the objective function.

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    $\begingroup$ The value of the minimum can be checked with GeoGebra to be about $6.4$. $\endgroup$
    – Intelligenti pauca
    Commented May 24 at 17:02
  • $\begingroup$ The answer has been updated to use the same objective function as given in the OP and agrees in result with that derived by Oth S. $\endgroup$
    – smichr
    Commented May 24 at 18:43
  • $\begingroup$ [+1] Good idea in particular to use half-angle formulas. $\endgroup$
    – Jean Marie
    Commented May 24 at 20:14
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    $\begingroup$ I am interested in the differences in approaches. I would consider the CAS a brute-force method. Conversion to tan(t/2) leads to a very intimidating equation which (if you can see it) factors into terms with the highest term having a quarter base to solve whereas the geometric approach below is so clean. I don't have the words to explain this difference about what is going on in each, but the fact that there are different paths to the answer requiring different levels of abilities is interesting in its own right. $\endgroup$
    – smichr
    Commented May 24 at 21:14
  • $\begingroup$ Thank you for sharing this detailed derivation using SymPy! It's a great demonstration of how computational tools can solve such optimization problems. The geometric method indeed provides an elegant and intuitive approach, while the computational method verifies the result through algebraic manipulation. $\endgroup$
    – Oth S
    Commented May 25 at 4:55
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Conceptually, an easy way to obtain the solution using tangency properties:

Given

$$ \cases{ \|p-p_a\|+\frac 12\|p-p_c\|=f\\ \|p-p_b\|^2 = 4\\ p_a = (1,0)\\ p_b = (5,0)\\ p_c = (0,5)} $$

at tangency, eliminating $y$ we get

$$ p(x) = f^8-48 f^6 x+4 f^6+904 f^4 x^2-1808 f^4 x+4366 f^4-4672 f^2 x^3+1744 f^2 x^2+39088 f^2 x-159276 f^2+26896 x^4-280768 x^3+1448104 x^2-3733872 x+4756761 $$

this polynomial should have a double root (due to tangency) or

$$ p(x) = k(x-x_1)^2(x^2+a x + b)\ \ \forall x $$

or equivalently

$$ \left\{ \begin{array}{rcl} -b k x_1^2+f^8+4 f^6+4366 f^4-159276 f^2+4756761 & = & 0\\ -a k x_1^2+2 b k x_1-48 f^6-1808 f^4+39088 f^2-3733872 & = & 0\\ 2 a k x_1-b k+904 f^4+1744 f^2-k x_1^2+1448104 & = & 0\\ -a k-4672 f^2+2 k x_1-280768 & = & 0\\ 26896-k & = & 0\\ \end{array} \right. $$

now solving for $\{f,k,a,b,x_1\}$ we obtain the solution.

Follows a MATHEMATICA scrip to solve the problem.

pa = {1, 0};
pb = {5, 0};
pc = {0, 5};
p = {x, y};
c1 = Sqrt[(p - pc) . (p - pc)] + 1/2 Sqrt[(p - pa) . (p - pa)] - f
c2 = (p - pb) . (p - pb) - 2^2
polx = First[Eliminate[{c1 == 0, c2 == 0}, y]] - k (x - x1)^2 (x^2 + a x + b)
Solve[CoefficientList[polx, x] == 0, {k, f, x1, a, b}, Reals]

gr1 = ContourPlot[Sqrt[(p - pc) . (p - pc)] + 1/2 Sqrt[(p - pa) . (p - pa)] == Sqrt[41], {x, -1, 8}, {y, -3, 6}];
gr2 = ContourPlot[(p - pb) . (p - pb) - 2^2 == 0, {x, -1, 8}, {y, -3, 6}];
points = {pa, pb, pc};
gr0 = Graphics[{Red, PointSize[0.02], Point[points]}];
Show[gr1, gr2, gr0]

enter image description here

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