327. Some geometrical/algebraic extremum problems.
by Virgil Nicula, Oct 31, 2011, 7:52 AM
PP1. Prove that in an isosceles triangle 
with 
following inequality holds 
, where 
is inradius.
Proof 1. Observe that
, where 
, 
. Prove (with derivatives) that for any 
we have
. But 
. Thus 
.
Remark. If the triangle
is no acute, then 
. This 
is a good idea and yet 
Proof 2.
and 
. We"ll show 
, i.e. 
for any 
(inequality of triangle : 
). Denote 
.
Obtain
for any 
. Let 
, i.e. 
. Obtain 
for any 
. It is truly because 
and the discriminant
of 
is 
(negative), i.e 
. So, 
, i.e. 
.
Remark. I like this problem. We can find a better approximation if consider the hexagon
which is circumscribed to the incircle and 
so that 
. Observe that 
and the semiperimeter of this hexagon is greater than the semiperimeter of the incircle, i.e. 
.
Directly,
.
PP2. (USAMTS 2009 Round 2 #5). Let
with the circumcircle 
and 
, 
. For 
denote 
. Prove that 
.
Proof. First we set up a coordinate system with
, 
, 
. Try to come up with an expression in 
for 
. First we find the polar equation for
the circumcircle of
, then find the polar equation for line 
, then subtract the two values of 
to get length . For the circumcircle of the rectangular equation is
. We then let 
and expand in 
.
Now to find the equation for line, the rectangular equation is 
, and again we have to plug in 
. We subtract
the second from the first :
. We know that
. We used inequality twice, once when we assumed 
and once when we assumed
. The first equality happens when 
and the second equality happens only when 
. Since they can't happen simultaneously,
the
is actually a 
sign. In the original problem, it was in fact 
, not 
.
Lemma.
, 
and 
.
Proof 1. For
is easily ! Indeed, can suppose w.l.o.g. that 
. Denote 
. Prove easily 
that ![$t\in \left[1,\sqrt 2\right]$](//latex.artofproblemsolving.com/4/f/b/4fb18a3248ba98daaa154ced66611f15d17a24d5.png)
. Thus, 
is min.
is max. 
is max. 
is max. The function 
is 
(strict increasing) on ![$\left[0,\sqrt 2\right]$](//latex.artofproblemsolving.com/6/0/8/608c2b3b57fffbac4abac8735cb35b51f6429a42.png)
. In conclusion, this function is max.
, i.e. 
. What happen if 
? This is the actual problem. Answer. ![$\frac ax+\frac by\ge \left(\sqrt[3]{a^2}+\sqrt [3]{b^2}\right)^{\frac 32}$](//latex.artofproblemsolving.com/6/d/0/6d05501347e6485d7393c29201acf1980444ce39.png)
with equality iff ![$\frac xy=\sqrt[3]{\frac ab}$](//latex.artofproblemsolving.com/c/c/b/ccb21c306fcf5e5f2e275c7d7b9ffeb3f374aca7.png)
.
Remark. 
.
Proof 2. Apply the Holder's inequality which works quite nicely in this case, indeed, we need to establish
the following
![$ \left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)^3$](//latex.artofproblemsolving.com/b/c/f/bcfe6e5315900479caa8049ad3fbbf87f7fcd89b.png)
.
Extension. Let
-right with circumcircle 
, 
. For denote
. Prove that is maximum ![$\iff \frac {PB}{PC}=\sqrt [3]{\left(\frac cb\right)^2}\iff \tan\widehat{PAB}=\sqrt[3]{\frac bc}$](//latex.artofproblemsolving.com/8/a/5/8a5cd9672961c123dee434f12e5f986f80edb50d.png)
.
Proof. Let
. From 
and the Sinus' theorem in 
.
So, is 
is 
is ![$\tan x=\sqrt [3]{\frac bc}\iff\frac {PB}{PC}=\frac cb\cdot\tan x=\sqrt [3]{\left(\frac cb\right)^2}$](//latex.artofproblemsolving.com/d/d/8/dd8a2708026920a9c649a17b103b1058ad008ab8.png)
. Thus,
. In the particular case and obtain 
.
PP3. Find minimum value of
.
Proof 1 (algebraic). Using
with equality when 
and 
with equality when 
obtain
that
![$2\left[\sqrt{\left(x-\frac{3}{2}\right)^2+1}+\sqrt{\left(x-\frac{7}{2}\right)^2+1}\right]\ge$](//latex.artofproblemsolving.com/e/9/b/e9b30372e2e8b93e6d785a334c66a8558044c059.png)
with equality when
and 
.
Proof 2 (geometric). Consider
, 
and a mobile point 
-axis. Denote symmetrical 
of w.r.t. 
-axis and 
.
Observe that
and 
is minimum 
is minimum 
. The equation of 
is 
and for
obtain that 
. In conclusion, 
and 
.
Proof 3. Apply to
the Minkowski's inequality 
.
Obtain that
with equality 
.
PP4. Let
be a convex pentagon inscribed in the circle 
and for which 
. Ascertain its maximum area ![$ [ABCDE]$](//latex.artofproblemsolving.com/9/6/4/964e1a67199afc81d2077cdee30359adbb71f9de.png)
.
Proof. Suppose w.l.o.g.
. Let 
and the distance 
of 
to 
. Since obtain 
and
. Apply Ptolemy's in 
: 
. So, ![$ [ABCDE] = [ABCD] + [AED] =$](//latex.artofproblemsolving.com/d/4/2/d4205040e1cf02d04819172941dfdd05a685724b.png)
. So, the area is 
![$ \left[2uv + y(2R + x)\right]$](//latex.artofproblemsolving.com/2/f/4/2f4d3ede802215b9197bb017c74e5abd10d229c9.png)
is 
with 
.
.
si 
. In conclusion, 
, 
and 
.
PP5. Prove that
.
Proof. Observe that
. Thus, 
. We have equality 
(never). In conclusion, 
.
PP6. A right circular cylindrical container with a closed top is to be constructed with a fixed surface area. Find the ratio of the height to the radius which will maximize the volume.
Proof. Let , 
be the radius and the height respectively. Then the surface area is 
(constant) and the volume is 
.
Our problem is "maximize the product
, where the sum 
(constant)". So, is 
is , where 
(constant). If the sum is constant, then the product is 
. (the axial section is a square).
PP7. Find the maximum value of
, where ![$x\in [1,6]$](//latex.artofproblemsolving.com/e/c/9/ec950cb85eefcc32413e97ca7103268b95ef75be.png)
.
Proof. Denote![$f(x)=x^{3}-9x^{2}+24x+5\ ,\ x\in [1,6]$](//latex.artofproblemsolving.com/1/4/e/14eb5a73fdeee430c03e09172712e9706322469f.png)
. Therefore, 
and 
Therefore,
. The range of 
, where ![$I=[1,6]$](//latex.artofproblemsolving.com/c/1/b/c1b3d5e82e2b674621a317372cfe6019db2a49aa.png)
is ![$\boxed{\mathrm{Im}(f)=f(I)=[21,41]}$](//latex.artofproblemsolving.com/f/6/4/f64272cd0b898a47f92a7a0f8a71d8acbb353cda.png)
.
Verify
PP8. Let
be two given positive numbers for which 
. Find the minimum value of 
.
Proof. There is
so that 
and 
. Consider the right and isosceles so that 
and 
, the fixed 
so that
and a mobile 
so that 
. Apply the generalized Pythagoras' theorem to the sides 
, 
in the trangles 
, 
respectively :
with equality iff 
, i.e. 
and
.
An easy extension. Let
be positive numbers so that 
. Find the minimum of 
.
Answer :
, 
.
PP9. Let
be constants and real numbers 
so 
. Prove that ![$\sqrt[3]{x^2+y^2}\ge \sqrt [3]{a^2}+\sqrt [3]{b^2}$](//latex.artofproblemsolving.com/0/9/a/09aa08f710482b2559f89e8b2605ac7939396c38.png)
with equality iff ![$\frac {x}{\sqrt[3]a}=\frac {y}{\sqrt [3]b}=\sqrt[3]{a^2}+\sqrt[3]{b^2}$](//latex.artofproblemsolving.com/6/c/2/6c27ea5d1ade292fefd916a828a5df506c979637.png)
.
A geometrical interpretation. Let an acute . For a mobile ![$M\in [AB]$](//latex.artofproblemsolving.com/5/5/6/556a215c7a0922d0160f9e6c380d92ffecc7f57e.png)
let 
. Find position of 
so distance 
is maximum.
Proof 1.
![$ \left(\sqrt[3]{\frac ax\cdot\frac ax\cdot x^2}+\sqrt[3]{\frac by\cdot\frac by\cdot y^2}\right)^3=$](//latex.artofproblemsolving.com/e/4/d/e4d09bafaf36d4bfda69fb8be9235b3e517072ce.png)
![$\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)^3$](//latex.artofproblemsolving.com/c/c/f/ccf586c4e98f9bc51d0d00fe954d651cf497c0e5.png)
.
Proof 2.
![$f'(x)\ .s.s.\ \left[(x-a)^3-ab^2\right]\ .s.s.\ \left(x-a-\sqrt[3]{ab^2}\right)$](//latex.artofproblemsolving.com/e/9/5/e95e101d213ec16895332811fcbaf48af9d1e03a.png)
.
Thus,![$\left\{\begin{array}{ccccc}
x_{\mathrm{min}} & = & a+\sqrt[3]{ab^2} & = & \sqrt[3]a\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)\\\\
y_{\mathrm{min}} & = & b+\sqrt[3]{a^2b} & = & \sqrt[3]b\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)\end{array}\right|$](//latex.artofproblemsolving.com/5/5/2/5523f5d361cedbc9fb8d787d57a2baa1c0038692.png)
and 
![$x_{\mathrm{min}}^2+y_{\mathrm{min}}^2=\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)^3$](//latex.artofproblemsolving.com/c/2/b/c2b45f35ddc20bbc6ded45ddaee6f32977a42238.png)
.
PP10. Let
and 
so that 
. Find the minimum value of 
, where 
.
Proof (without derivatives).
. Indeed, 
. Have equality iff 
, i.e. 
, where ![$t_0=\sqrt {(a-1)(b-1)}-1\in [0,1]$](//latex.artofproblemsolving.com/8/8/e/88e5da1cc11a6889f29ca4d0e9006ee6fcd53636.png)
.
PP11. Let
be a square with 
and 
is a circle with ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
as its diameter. The segment ![$[PQ]$](//latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
with 
, 
and 
intersects . Let 
be the area in the square below and outside 
and let 
be the area above and inside . How far should be from to minimize 
?
Proof 1. Let
, where ![$x\in\left[0,\frac 12\right]$](//latex.artofproblemsolving.com/2/8/2/2821378dbb628d2e1d61f9ee01273aee884168ca.png)
and area 
of common surface between square and . Prove easily 
. Thus, 
is minimum 
. For 
and 
![$\cos\phi =2x\ ,\ [MON]=$](//latex.artofproblemsolving.com/5/6/a/56a4698880b45d3cc812f6ade326ecd309cd47ae.png)
![$\boxed{[MON]=\frac {x\sqrt{1-4x^2}}{2}}$](//latex.artofproblemsolving.com/8/a/7/8a72a8d3e27589bcb3e98573835cafc334eb114f.png)
.
The length of
is equally to 
and area of the sector 
is equally to 
. Thus, 
is maximum
is maximum. So 
and 
. Hence 
is minimum ![$\boxed{x_{\mathrm{min}}=\frac {\sqrt 3}{4}\in\left[0,\frac 12\right]}$](//latex.artofproblemsolving.com/8/4/9/849cf70624efa34b946b44db4af472fcec6c69aa.png)
, i.e. 
and in this case 
.
Proof 2. Denote the area of the common surface between square and circle , the intersection 
and 
, where ![$\phi\in\left[0,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/8/7/6/876cea07f0de8be76457a92a1a237e10d79f07ff.png)
.
Prove easily that
, 
, ![$[MON]=\frac 18\cdot\sin 2\phi$](//latex.artofproblemsolving.com/0/b/2/0b24df1e321b8bc6b7e72a195a035e8efb93562b.png)
, the length of the little arc 
is equally to 
and the area of the sector 
is equally to 
. So
and 
. Thus, 
is minimum 
. Thus, 
.
In conclusion, is minimum ![$\boxed{\phi_{\mathrm{min}}=\frac {\pi}{6}\in\left[0,\frac {\pi}{2}\right]}$](//latex.artofproblemsolving.com/b/b/9/bb9a2140fa49a9d161521d6c64bd5ab32eb26f16.png)
, i.e. .
PP12. Let and , 
so that . Know 
and 
. Find the range of 
.
Proof. Suppose w.l.o.g.
and denote 
. Consider the parallelogram 
, where 
, 
and 
. Thus,
. Observe that 
and 
. Apply the
theorem of Sines in
.
In conclusion,
, where .
PP13. Let
which satisfy 
, where 
. Find 
for which the difference between maximum minimum possible values of 
equals 
.
Proof. Let
. Thus, 
and 
. So, 
whose discriminant is 
. Then 
and
. So we are looking for 
and so 
.
with 
following inequality holds 
, where 
is inradius.Proof 1. Observe that
, where 
, 
. Prove (with derivatives) that for any 
we have
. But 
. Thus 
.Remark. If the triangle
is no acute, then 
. This 
is a good idea and yet 
Proof 2.
and 
. We"ll show 
, i.e. 
for any 
(inequality of triangle : 
). Denote 
.Obtain
for any 
. Let 
, i.e. 
. Obtain 
for any 
. It is truly because 
and the discriminant
of 
is 
(negative), i.e 
. So, 
, i.e. 
.Remark. I like this problem. We can find a better approximation if consider the hexagon
which is circumscribed to the incircle and 
so that 
. Observe that 
and the semiperimeter of this hexagon is greater than the semiperimeter of the incircle, i.e. 
.Directly,
. PP2. (USAMTS 2009 Round 2 #5). Let
with the circumcircle 
and 
, 
. For 
denote 
. Prove that 
. Proof. First we set up a coordinate system with
, 
, 
. Try to come up with an expression in 
for 
. First we find the polar equation forthe circumcircle of
, then find the polar equation for line 
, then subtract the two values of 
to get length . For the circumcircle of the rectangular equation is
. We then let 
and expand in 
.Now to find the equation for line
, the rectangular equation is 
, and again we have to plug in 
. We subtractthe second from the first :
. We know that
. We used inequality twice, once when we assumed 
and once when we assumed
. The first equality happens when 
and the second equality happens only when 
. Since they can't happen simultaneously,the
is actually a 
sign. In the original problem, it was in fact 
, not 
.Lemma.
, 
and 
.Proof 1. For
is easily ! Indeed, can suppose w.l.o.g. that 
. Denote 
. Prove easily 
that ![$t\in \left[1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/4/f/b/4fb18a3248ba98daaa154ced66611f15d17a24d5.png)
. Thus, 
is min. 
is max. 
is max. 
is max. The function 
is 
(strict increasing) on ![$\left[0,\sqrt 2\right]$](http://latex.artofproblemsolving.com/6/0/8/608c2b3b57fffbac4abac8735cb35b51f6429a42.png)
. In conclusion, this function is max. 
, i.e. 
. What happen if 
? This is the actual problem. Answer. ![$\frac ax+\frac by\ge \left(\sqrt[3]{a^2}+\sqrt [3]{b^2}\right)^{\frac 32}$](http://latex.artofproblemsolving.com/6/d/0/6d05501347e6485d7393c29201acf1980444ce39.png)
with equality iff ![$\frac xy=\sqrt[3]{\frac ab}$](http://latex.artofproblemsolving.com/c/c/b/ccb21c306fcf5e5f2e275c7d7b9ffeb3f374aca7.png)
. Remark. 
.Proof 2. Apply the Holder's inequality which works quite nicely in this case, indeed, we need to establish
the following
![$ \left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)^3$](http://latex.artofproblemsolving.com/b/c/f/bcfe6e5315900479caa8049ad3fbbf87f7fcd89b.png)
.Extension. Let
-right with circumcircle 
, 
. For denote . Prove that is maximum ![$\iff \frac {PB}{PC}=\sqrt [3]{\left(\frac cb\right)^2}\iff \tan\widehat{PAB}=\sqrt[3]{\frac bc}$](http://latex.artofproblemsolving.com/8/a/5/8a5cd9672961c123dee434f12e5f986f80edb50d.png)
. Proof. Let
. From 
and the Sinus' theorem in 
.So,
is 
is 
is ![$\tan x=\sqrt [3]{\frac bc}\iff\frac {PB}{PC}=\frac cb\cdot\tan x=\sqrt [3]{\left(\frac cb\right)^2}$](http://latex.artofproblemsolving.com/d/d/8/dd8a2708026920a9c649a17b103b1058ad008ab8.png)
. Thus,
. In the particular case and obtain 
.PP3. Find minimum value of
.Proof 1 (algebraic). Using
with equality when 
and 
with equality when 
obtainthat
![$2\left[\sqrt{\left(x-\frac{3}{2}\right)^2+1}+\sqrt{\left(x-\frac{7}{2}\right)^2+1}\right]\ge$](http://latex.artofproblemsolving.com/e/9/b/e9b30372e2e8b93e6d785a334c66a8558044c059.png)
with equality when
and 
.Proof 2 (geometric). Consider
, 
and a mobile point 
-axis. Denote symmetrical 
of w.r.t. 
-axis and 
.Observe that
and 
is minimum 
is minimum 
. The equation of 
is 
and for
obtain that 
. In conclusion, 
and 
.Proof 3. Apply to
the Minkowski's inequality 
.Obtain that
with equality 
.PP4. Let
be a convex pentagon inscribed in the circle 
and for which 
. Ascertain its maximum area ![$ [ABCDE]$](http://latex.artofproblemsolving.com/9/6/4/964e1a67199afc81d2077cdee30359adbb71f9de.png)
.Proof. Suppose w.l.o.g.
. Let 
and the distance 
of 
to 
. Since obtain 
and
. Apply Ptolemy's in 
: 
. So, ![$ [ABCDE] = [ABCD] + [AED] =$](http://latex.artofproblemsolving.com/d/4/2/d4205040e1cf02d04819172941dfdd05a685724b.png)
. So, the area is 
![$ \left[2uv + y(2R + x)\right]$](http://latex.artofproblemsolving.com/2/f/4/2f4d3ede802215b9197bb017c74e5abd10d229c9.png)
is 
with 
.
.
si 
. In conclusion, 
, 
and 
.PP5. Prove that
.Proof. Observe that
. Thus, 
. We have equality 
(never). In conclusion, 
.PP6. A right circular cylindrical container with a closed top is to be constructed with a fixed surface area. Find the ratio of the height to the radius which will maximize the volume.
Proof. Let
, 
be the radius and the height respectively. Then the surface area is 
(constant) and the volume is 
.Our problem is "maximize the product
, where the sum 
(constant)". So, is 
is , where 
(constant). If the sum is constant, then the product is 
. (the axial section is a square).PP7. Find the maximum value of
, where ![$x\in [1,6]$](http://latex.artofproblemsolving.com/e/c/9/ec950cb85eefcc32413e97ca7103268b95ef75be.png)
.Proof. Denote
![$f(x)=x^{3}-9x^{2}+24x+5\ ,\ x\in [1,6]$](http://latex.artofproblemsolving.com/1/4/e/14eb5a73fdeee430c03e09172712e9706322469f.png)
. Therefore, 
and 
Therefore,
. The range of 
, where ![$I=[1,6]$](http://latex.artofproblemsolving.com/c/1/b/c1b3d5e82e2b674621a317372cfe6019db2a49aa.png)
is ![$\boxed{\mathrm{Im}(f)=f(I)=[21,41]}$](http://latex.artofproblemsolving.com/f/6/4/f64272cd0b898a47f92a7a0f8a71d8acbb353cda.png)
.Verify
PP8. Let
be two given positive numbers for which 
. Find the minimum value of 
.Proof. There is
so that 
and 
. Consider the right and isosceles so that 
and 
, the fixed 
so that
and a mobile 
so that 
. Apply the generalized Pythagoras' theorem to the sides 
, 
in the trangles 
, 
respectively :
with equality iff 
, i.e. 
and
.An easy extension. Let
be positive numbers so that 
. Find the minimum of 
.Answer :
, 
.PP9. Let
be constants and real numbers 
so 
. Prove that ![$\sqrt[3]{x^2+y^2}\ge \sqrt [3]{a^2}+\sqrt [3]{b^2}$](http://latex.artofproblemsolving.com/0/9/a/09aa08f710482b2559f89e8b2605ac7939396c38.png)
with equality iff ![$\frac {x}{\sqrt[3]a}=\frac {y}{\sqrt [3]b}=\sqrt[3]{a^2}+\sqrt[3]{b^2}$](http://latex.artofproblemsolving.com/6/c/2/6c27ea5d1ade292fefd916a828a5df506c979637.png)
.A geometrical interpretation. Let an acute
. For a mobile ![$M\in [AB]$](http://latex.artofproblemsolving.com/5/5/6/556a215c7a0922d0160f9e6c380d92ffecc7f57e.png)
let 
. Find position of 
so distance 
is maximum.Proof 1.
![$ \left(\sqrt[3]{\frac ax\cdot\frac ax\cdot x^2}+\sqrt[3]{\frac by\cdot\frac by\cdot y^2}\right)^3=$](http://latex.artofproblemsolving.com/e/4/d/e4d09bafaf36d4bfda69fb8be9235b3e517072ce.png)
![$\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)^3$](http://latex.artofproblemsolving.com/c/c/f/ccf586c4e98f9bc51d0d00fe954d651cf497c0e5.png)
.Proof 2.
![$f'(x)\ .s.s.\ \left[(x-a)^3-ab^2\right]\ .s.s.\ \left(x-a-\sqrt[3]{ab^2}\right)$](http://latex.artofproblemsolving.com/e/9/5/e95e101d213ec16895332811fcbaf48af9d1e03a.png)
.Thus,
![$\left\{\begin{array}{ccccc}
x_{\mathrm{min}} & = & a+\sqrt[3]{ab^2} & = & \sqrt[3]a\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)\\\\
y_{\mathrm{min}} & = & b+\sqrt[3]{a^2b} & = & \sqrt[3]b\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)\end{array}\right|$](http://latex.artofproblemsolving.com/5/5/2/5523f5d361cedbc9fb8d787d57a2baa1c0038692.png)
and 
![$x_{\mathrm{min}}^2+y_{\mathrm{min}}^2=\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)^3$](http://latex.artofproblemsolving.com/c/2/b/c2b45f35ddc20bbc6ded45ddaee6f32977a42238.png)
.PP10. Let
and 
so that 
. Find the minimum value of 
, where 
.Proof (without derivatives).
. Indeed, 
. Have equality iff 
, i.e. 
, where ![$t_0=\sqrt {(a-1)(b-1)}-1\in [0,1]$](http://latex.artofproblemsolving.com/8/8/e/88e5da1cc11a6889f29ca4d0e9006ee6fcd53636.png)
.PP11. Let
be a square with 
and 
is a circle with ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
as its diameter. The segment ![$[PQ]$](http://latex.artofproblemsolving.com/2/1/c/21ca08816cf8b23ddf756ce9ae098ad327f2443d.png)
with 
, 
and 
intersects
. Let 
be the area in the square below and outside 
and let 
be the area above and inside . How far should be from to minimize 
?Proof 1. Let
, where ![$x\in\left[0,\frac 12\right]$](http://latex.artofproblemsolving.com/2/8/2/2821378dbb628d2e1d61f9ee01273aee884168ca.png)
and area 
of common surface between square and . Prove easily 
. Thus, 
is minimum
. For 
and 
![$\cos\phi =2x\ ,\ [MON]=$](http://latex.artofproblemsolving.com/5/6/a/56a4698880b45d3cc812f6ade326ecd309cd47ae.png)
![$\boxed{[MON]=\frac {x\sqrt{1-4x^2}}{2}}$](http://latex.artofproblemsolving.com/8/a/7/8a72a8d3e27589bcb3e98573835cafc334eb114f.png)
.The length of
is equally to 
and area of the sector 
is equally to 
. Thus, 
is maximum 
is maximum. So 
and 
. Hence 
is minimum ![$\boxed{x_{\mathrm{min}}=\frac {\sqrt 3}{4}\in\left[0,\frac 12\right]}$](http://latex.artofproblemsolving.com/8/4/9/849cf70624efa34b946b44db4af472fcec6c69aa.png)
, i.e. 
and in this case 
.Proof 2. Denote the area
of the common surface between square and circle , the intersection 
and 
, where ![$\phi\in\left[0,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/8/7/6/876cea07f0de8be76457a92a1a237e10d79f07ff.png)
.Prove easily that
, 
, ![$[MON]=\frac 18\cdot\sin 2\phi$](http://latex.artofproblemsolving.com/0/b/2/0b24df1e321b8bc6b7e72a195a035e8efb93562b.png)
, the length of the little arc 
is equally to 
and the area of the sector 
is equally to 
. So
and 
. Thus, 
is minimum 
. Thus, 
.In conclusion,
is minimum ![$\boxed{\phi_{\mathrm{min}}=\frac {\pi}{6}\in\left[0,\frac {\pi}{2}\right]}$](http://latex.artofproblemsolving.com/b/b/9/bb9a2140fa49a9d161521d6c64bd5ab32eb26f16.png)
, i.e. .PP12. Let
and , 
so that . Know 
and 
. Find the range of 
.Proof. Suppose w.l.o.g.
and denote 
. Consider the parallelogram 
, where 
, 
and 
. Thus,
. Observe that 
and 
. Apply thetheorem of Sines in
.In conclusion,
, where .PP13. Let
which satisfy 
, where 
. Find 
for which the difference between maximum minimum possible values of 
equals 
.Proof. Let
. Thus, 
and 
. So, 
whose discriminant is 
. Then 
and
. So we are looking for 
and so 
.This post has been edited 169 times. Last edited by Virgil Nicula, Jan 5, 2016, 8:45 AM
August 2018
January 2018