417. The incircle of ABC and other two incircles.
by Virgil Nicula, Mar 23, 2015, 8:15 PM
PP1 (Ruben Auqui). Let 
be an 
-right triangle with the incircle 
. For 
denote the incircle 
of
and the incircle 
of 
. Prove that 
.
Proof. Denote the tangent points
and 
, where 
.
Observe that
. Therefore, 
![$[2\rho +d-(b+c-a)]^2=4\rho^2+d^2\iff$](//latex.artofproblemsolving.com/0/8/3/083dcec942e8941d700d4b7dce9ded0fc9cca699.png)
![$\left[2\rho +d-2(s-a)\right]^2=$](//latex.artofproblemsolving.com/5/c/0/5c08c1de72749ce9c36ca9a020ac7bd95749ffe0.png)
![$4\rho^2+d^2\stackrel{s-a=r}{\iff}\ [d-2(r-\rho )]^2=d^2+4\rho^2\iff$](//latex.artofproblemsolving.com/7/1/5/715b9cf2098717f7cd5a32656340dee9de4bf48b.png)
. In the particular case 
and 
obtain that 
.
PP2. Let a
-isosceles 
with 
so that 
and 
. Prove that ![$\max [ABC]=50$](//latex.artofproblemsolving.com/6/d/0/6d043ef01e7facb049cf14f518de17f4d1846620.png)
.
Proof 1. Denote
, the midpoint 
of ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and 
. Apply the Menelaus's theorem to the transversal 
. Thus, ![$\frac {[ARM]}{[ABC]}=\frac {[ARM]}{[ABM]}\cdot\frac{[ABM]}{[ABC]}=$](//latex.artofproblemsolving.com/f/4/2/f42a822a336102ce6ac5b8f942d4e3d8db9a4d94.png)
![$[ABC]=8\cdot [ARM]$](//latex.artofproblemsolving.com/3/0/e/30e99a23555a2642e806a4d828a6d8cebe379a6c.png)
and 
and 
. So, ![$[ABC]$](//latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
is 
![$[ARM]$](//latex.artofproblemsolving.com/f/8/e/f8ec8500a9875787983ef743b1b4891a84ecab0c.png)
is 
is 
is
is 
, i.e. 
because the sum 
(constant). Thus, ![$[ABC]_{\mathrm{max}}=8\cdot \frac 12\cdot x^2=4x^2=50$](//latex.artofproblemsolving.com/2/1/1/2113b8601470df55da8ef78fe983b0634487f042.png)
.
Proof 2.![$[ABC]=\frac 53\cdot[ABP]$](//latex.artofproblemsolving.com/d/e/a/dea787bc46d9c990bd5e3ee5fc08f9a4d9b66aea.png)
is ![$[ABP]$](//latex.artofproblemsolving.com/4/8/8/48884bc091512a29bd03f48e81ba55152b010e59.png)
is 
, where and 
. Heron's ![$:\ [ABP]=\sqrt{4(k+1)\cdot (4-k)\cdot (k+4)\cdot 4(k-1)}\implies$](//latex.artofproblemsolving.com/5/2/0/5206b090991f993546a5b8f65af7ec6eff568959.png)
![$\boxed{[ABP]=4\sqrt{\left(k^2-1\right)\left(16-k^2\right)}}\ (*)$](//latex.artofproblemsolving.com/d/6/1/d61a2724374057056fd2c549edd56a90b2c12fad.png)
. In conclusion, is 
![$[ABC]_{\mathrm{max}}\stackrel{(*)}{=}\frac 53\cdot 4\cdot\left(16-\frac {17}2\right)\implies$](//latex.artofproblemsolving.com/b/8/6/b869466c6e90c71523f55667d1965f0b0563da6a.png)
![$[ABC]_{\mathrm{max}}=50$](//latex.artofproblemsolving.com/e/8/1/e81e84ba3725c7318cc3fa82987891738ef1284a.png)
.
Proof 3.![$\frac {BC}{BP}=\frac 53\implies 3\cdot [ABC]=5\cdot [ABP]$](//latex.artofproblemsolving.com/1/e/e/1eef5a8215a2769469714b31f1cbfae9193dda8c.png)
. Thus, is is , where and 
. Let the points 
so that 
and
, i.e. the division 
is harmonically. Observe that 
and 
. Let the Appolonius's circle 
with the diameter ![$[DE]$](//latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
. Show easily that 
and is 
and 
, i.e. 
. The maximum area of is 
![$[ABP]_{\mathrm{max}}=30\iff$](//latex.artofproblemsolving.com/e/9/e/e9e412fe0d924ab0bbfe68fc3818943b6781103f.png)
![$[ABC]_{\mathrm{max}}=\frac 53\cdot [ABP]_{\mathrm{max}}=50$](//latex.artofproblemsolving.com/a/4/5/a45f1a7e0791a93f24ad6ff6758804f17f2b0f03.png)
.
Proof 4.
, and 
. Apply the Stewart's teorem in 
. The area ![$\boxed{[ABC]=\frac 12\cdot x\cdot\sqrt{25k^2-\left(\frac x2\right)^2}}\ (1)$](//latex.artofproblemsolving.com/5/0/c/50c6ee49225fe7f3ed34617fe1c8d4eb2efaa34d.png)
is 
is 
![$x^2\left[5\left(320-3x^2\right)-x^2\right]$](//latex.artofproblemsolving.com/a/b/2/ab21608987d2be3b1ee9afa9528391e0f522e862.png)
is 
is 
and 
. From 
obtain easily that ![$[ABC]_{\max}=50$](//latex.artofproblemsolving.com/3/2/9/32959526a89648dd2fc3bb28042cbceff110a7d8.png)
The Descartes circle theorem. Given three noncollinear points, construct three tangent circles such that one is centered at each point and the circles are
pairwise tangent to one another. Then there exist exactly two nonintersecting circles that are tangent to all three circles. These are called the inner and
outer Soddy circles. Frederick Soddy (1936) gave the formula for finding the radii
of the Soddy circles given the radii 
of the other three 
With other words, 
.
Here, the negative solution corresponds to the outer Soddy circle and the positive one to the inner Soddy circle. Here, the negative solution corresponds to
the outer Soddy circle and the positive one to the inner Soddy circle. This formula is called the Descartes circle theorem since it was known to Descartes.
PP3. Let
be two circles what are exterior tangent at the point 
and for which denote a common exterior tangent 
, where 
. Let the circle 
which is tangent to the line 
and is exterior tangent to the circles 
so that its center 
is an interior point of 
. Let
the circle
what is exterior tangent to the circles 
and . Prove that 
and 
.
Proof. Apply the Descartes' theorem to the circles
, 
and with the radius 
(degenerate) 
.
Denote
![$\left\{\begin{array}{cccc}
x^2y^2+z^2\left(x^2+y^2\right)=x^2y^2+z^2\left[(x+y)^2-2xy\right] & \stackrel{(1)}{=} & x^2y^2+x^2y^2-2xyz^2=2xy\left(xy-z^2\right) & (2)\\\\
z^2\left(x^2+y^2+z^2\right)=z^2\left[(x+y)^2-2xy+z^2\right] & \stackrel{(1)}{=} & x^2y^2-2xyz^2+z^4=\left(xy-z^2\right)^2 & (3)\end{array}\right\|$](//latex.artofproblemsolving.com/5/7/9/5799dab46d6274dfacf3b8984a0f26ecacc4a04d.png)
.Apply the
Descartes' theorem to
and 
![$\frac {x^2y^2z^2}{x^2y^2+z^2\left(x^2+y^2\right)+2\sqrt{x^2y^2z^2\left[\left(x^2+y^2\right)+z^2\right]}}\ \stackrel{(1\wedge 2)}{\implies}$](//latex.artofproblemsolving.com/8/9/9/89971f7a2195a5323b6bf5ff6498de466906e44f.png)
.
Remark. Otherwise. The line is the outer tangent circle (degenerate) to , and . Therefore, 
and the
relation
becomes 
PP4. Let and its interior 
for which there are 
so that 
. Prove that 
.
Proof. Let
. Prove easily that 
and ![$\cos\frac {\alpha}2=\sqrt{\frac {r[(r+(y+z)]}{(r+y)(r+z)}}$](//latex.artofproblemsolving.com/7/8/0/7805980e64056f287923ff256b2c968ecc182002.png)
a.s.o. I"ll apply an well-known conditioned identity
.Thus, 
![$-4r\sqrt{r\prod(r+y+z)}=(r+x)(r+y)(r+z)+\sum (r+x)[r(r+y+z)-yz]=$](//latex.artofproblemsolving.com/b/e/2/be2ca000325beea0ba76d8ec46ff43185cb7e9dd.png)
![$4r^3\prod[r+(y+z)]\implies$](//latex.artofproblemsolving.com/c/1/e/c1e2edeeb95b2a407a6ae683059d28b79c07720c.png)
.
PP5. Let and for which exist 
so 
. Prove that 
.
Lemma. Let for which 
. Denote the midpoint of ![$[BC]\ ;$](//latex.artofproblemsolving.com/2/c/e/2cee46ce04305480ad31681565572a638aaf48b5.png)
the points 
where its incircle 
touches
respectively 
. Then 
and the division 
is harmonically.
Proof. Apply an well known relation
. Let 
Thus,
, 
and 
From and 
results that 
, i.e. 
what means
Have immediatelly and the second part of the conclusion, i.e. 
is the polar of the point w.r.t. 
PP6. Let with its incircle , the midpoint of 
and 
. The lines passing through 
, 
,
and are parallelly to intersect again the incircle in 
and 
. The lines 
and 
cut at and 
. Prove that 
.
Proof Let
for which 
. From the product of the relations 
, 
, 
, 
(from the above
lemma: the division
is harmonically, i.e. the point 
coincides with the point ) we obtain 
, i.e. 
PP7. Let an acute with incircle 
and circumcircle 
. The circles 
are tangent internally to 
in the same 
.
The circle
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
is the semiperimeter and ![$ S=[ABC]$](//latex.artofproblemsolving.com/a/c/5/ac5489c655a5d312c5c91c4f77c55eadde984421.png)
is the area for 
.
Remark. Prove easily
The relations 
are well-known.
Proof I. Let
and Stewart's in 
.
Proof II. Apply the Pythagoras' theorem in the triangles :
Therefore,
, i.e. 
. Observe that 
, where 
is the - exinradius of .
Remark. Prove easily that two more interesting relations :
and 
.
be an 
-right triangle with the incircle 
. For 
denote the incircle 
of
and the incircle 
of 
. Prove that 
.Proof. Denote the tangent points
and 
, where 
.Observe that
. Therefore, 
![$[2\rho +d-(b+c-a)]^2=4\rho^2+d^2\iff$](http://latex.artofproblemsolving.com/0/8/3/083dcec942e8941d700d4b7dce9ded0fc9cca699.png)
![$\left[2\rho +d-2(s-a)\right]^2=$](http://latex.artofproblemsolving.com/5/c/0/5c08c1de72749ce9c36ca9a020ac7bd95749ffe0.png)
![$4\rho^2+d^2\stackrel{s-a=r}{\iff}\ [d-2(r-\rho )]^2=d^2+4\rho^2\iff$](http://latex.artofproblemsolving.com/7/1/5/715b9cf2098717f7cd5a32656340dee9de4bf48b.png)
. In the particular case 
and 
obtain that 
. PP2. Let a
-isosceles 
with 
so that 
and 
. Prove that ![$\max [ABC]=50$](http://latex.artofproblemsolving.com/6/d/0/6d043ef01e7facb049cf14f518de17f4d1846620.png)
.Proof 1. Denote
, the midpoint 
of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and 
. Apply the Menelaus's theorem to the transversal 
. Thus, ![$\frac {[ARM]}{[ABC]}=\frac {[ARM]}{[ABM]}\cdot\frac{[ABM]}{[ABC]}=$](http://latex.artofproblemsolving.com/f/4/2/f42a822a336102ce6ac5b8f942d4e3d8db9a4d94.png)
![$[ABC]=8\cdot [ARM]$](http://latex.artofproblemsolving.com/3/0/e/30e99a23555a2642e806a4d828a6d8cebe379a6c.png)
and 
and 
. So, ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
is 
![$[ARM]$](http://latex.artofproblemsolving.com/f/8/e/f8ec8500a9875787983ef743b1b4891a84ecab0c.png)
is 
is 
is 
is 
, i.e. 
because the sum 
(constant). Thus, ![$[ABC]_{\mathrm{max}}=8\cdot \frac 12\cdot x^2=4x^2=50$](http://latex.artofproblemsolving.com/2/1/1/2113b8601470df55da8ef78fe983b0634487f042.png)
.Proof 2.
![$[ABC]=\frac 53\cdot[ABP]$](http://latex.artofproblemsolving.com/d/e/a/dea787bc46d9c990bd5e3ee5fc08f9a4d9b66aea.png)
is ![$[ABP]$](http://latex.artofproblemsolving.com/4/8/8/48884bc091512a29bd03f48e81ba55152b010e59.png)
is 
, where and 
. Heron's ![$:\ [ABP]=\sqrt{4(k+1)\cdot (4-k)\cdot (k+4)\cdot 4(k-1)}\implies$](http://latex.artofproblemsolving.com/5/2/0/5206b090991f993546a5b8f65af7ec6eff568959.png)
![$\boxed{[ABP]=4\sqrt{\left(k^2-1\right)\left(16-k^2\right)}}\ (*)$](http://latex.artofproblemsolving.com/d/6/1/d61a2724374057056fd2c549edd56a90b2c12fad.png)
. In conclusion, is 
![$[ABC]_{\mathrm{max}}\stackrel{(*)}{=}\frac 53\cdot 4\cdot\left(16-\frac {17}2\right)\implies$](http://latex.artofproblemsolving.com/b/8/6/b869466c6e90c71523f55667d1965f0b0563da6a.png)
![$[ABC]_{\mathrm{max}}=50$](http://latex.artofproblemsolving.com/e/8/1/e81e84ba3725c7318cc3fa82987891738ef1284a.png)
.Proof 3.
![$\frac {BC}{BP}=\frac 53\implies 3\cdot [ABC]=5\cdot [ABP]$](http://latex.artofproblemsolving.com/1/e/e/1eef5a8215a2769469714b31f1cbfae9193dda8c.png)
. Thus, is is , where and 
. Let the points 
so that 
and
, i.e. the division 
is harmonically. Observe that 
and 
. Let the Appolonius's circle 
with the diameter ![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
. Show easily that 
and is 
and 
, i.e. 
. The maximum area of is 
![$[ABP]_{\mathrm{max}}=30\iff$](http://latex.artofproblemsolving.com/e/9/e/e9e412fe0d924ab0bbfe68fc3818943b6781103f.png)
![$[ABC]_{\mathrm{max}}=\frac 53\cdot [ABP]_{\mathrm{max}}=50$](http://latex.artofproblemsolving.com/a/4/5/a45f1a7e0791a93f24ad6ff6758804f17f2b0f03.png)
.Proof 4.
, and 
. Apply the Stewart's teorem in 
. The area ![$\boxed{[ABC]=\frac 12\cdot x\cdot\sqrt{25k^2-\left(\frac x2\right)^2}}\ (1)$](http://latex.artofproblemsolving.com/5/0/c/50c6ee49225fe7f3ed34617fe1c8d4eb2efaa34d.png)
is 
is 
![$x^2\left[5\left(320-3x^2\right)-x^2\right]$](http://latex.artofproblemsolving.com/a/b/2/ab21608987d2be3b1ee9afa9528391e0f522e862.png)
is 
is 
and 
. From 
obtain easily that ![$[ABC]_{\max}=50$](http://latex.artofproblemsolving.com/3/2/9/32959526a89648dd2fc3bb28042cbceff110a7d8.png)
The Descartes circle theorem. Given three noncollinear points, construct three tangent circles such that one is centered at each point and the circles are
pairwise tangent to one another. Then there exist exactly two nonintersecting circles that are tangent to all three circles. These are called the inner and
outer Soddy circles. Frederick Soddy (1936) gave the formula for finding the radii
of the Soddy circles given the radii 
of the other three 
With other words, 
.Here, the negative solution corresponds to the outer Soddy circle and the positive one to the inner Soddy circle. Here, the negative solution corresponds to
the outer Soddy circle and the positive one to the inner Soddy circle. This formula is called the Descartes circle theorem since it was known to Descartes.
PP3. Let
be two circles what are exterior tangent at the point 
and for which denote a common exterior tangent 
, where 
. Let the circle 
which is tangent to the line 
and is exterior tangent to the circles 
so that its center 
is an interior point of 
. Letthe circle
what is exterior tangent to the circles 
and . Prove that 
and 
. Proof. Apply the Descartes' theorem to the circles
, 
and with the radius 
(degenerate) 
.Denote
![$\left\{\begin{array}{cccc}
x^2y^2+z^2\left(x^2+y^2\right)=x^2y^2+z^2\left[(x+y)^2-2xy\right] & \stackrel{(1)}{=} & x^2y^2+x^2y^2-2xyz^2=2xy\left(xy-z^2\right) & (2)\\\\
z^2\left(x^2+y^2+z^2\right)=z^2\left[(x+y)^2-2xy+z^2\right] & \stackrel{(1)}{=} & x^2y^2-2xyz^2+z^4=\left(xy-z^2\right)^2 & (3)\end{array}\right\|$](http://latex.artofproblemsolving.com/5/7/9/5799dab46d6274dfacf3b8984a0f26ecacc4a04d.png)
.Apply theDescartes' theorem to
and 
![$\frac {x^2y^2z^2}{x^2y^2+z^2\left(x^2+y^2\right)+2\sqrt{x^2y^2z^2\left[\left(x^2+y^2\right)+z^2\right]}}\ \stackrel{(1\wedge 2)}{\implies}$](http://latex.artofproblemsolving.com/8/9/9/89971f7a2195a5323b6bf5ff6498de466906e44f.png)
.Remark. Otherwise. The line
is the outer tangent circle (degenerate) to , and . Therefore, 
and therelation
becomes 
PP4. Let
and its interior 
for which there are 
so that 
. Prove that 
.Proof. Let
. Prove easily that 
and ![$\cos\frac {\alpha}2=\sqrt{\frac {r[(r+(y+z)]}{(r+y)(r+z)}}$](http://latex.artofproblemsolving.com/7/8/0/7805980e64056f287923ff256b2c968ecc182002.png)
a.s.o. I"ll apply an well-known conditioned identity
.Thus, 
![$-4r\sqrt{r\prod(r+y+z)}=(r+x)(r+y)(r+z)+\sum (r+x)[r(r+y+z)-yz]=$](http://latex.artofproblemsolving.com/b/e/2/be2ca000325beea0ba76d8ec46ff43185cb7e9dd.png)
![$4r^3\prod[r+(y+z)]\implies$](http://latex.artofproblemsolving.com/c/1/e/c1e2edeeb95b2a407a6ae683059d28b79c07720c.png)
.PP5. Let
and for which exist 
so 
. Prove that 
.Lemma. Let
for which 
. Denote the midpoint of ![$[BC]\ ;$](http://latex.artofproblemsolving.com/2/c/e/2cee46ce04305480ad31681565572a638aaf48b5.png)
the points 
where its incircle 
touches
respectively 
. Then 
and the division 
is harmonically.Proof. Apply an well known relation
. Let 
Thus,
, 
and 
From and 
results that 
, i.e. 
what means
Have immediatelly and the second part of the conclusion, i.e. 
is the polar of the point w.r.t. 
PP6. Let
with its incircle , the midpoint of 
and 
. The lines passing through 
, 
,and are parallelly to
intersect again the incircle in 
and 
. The lines 
and 
cut at and 
. Prove that 
.Proof Let
for which 
. From the product of the relations 
, 
, 
, 
(from the abovelemma: the division
is harmonically, i.e. the point 
coincides with the point ) we obtain 
, i.e. 
PP7. Let an acute
with incircle 
and circumcircle 
. The circles 
are tangent internally to 
in the same 
.The circle
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
is the semiperimeter and ![$ S=[ABC]$](http://latex.artofproblemsolving.com/a/c/5/ac5489c655a5d312c5c91c4f77c55eadde984421.png)
is the area for 
.Remark. Prove easily
The relations 
are well-known.Proof I. Let
and Stewart's in 
.Proof II. Apply the Pythagoras' theorem in the triangles :
Therefore,
, i.e. 
. Observe that 
, where 
is the - exinradius of .Remark. Prove easily that two more interesting relations :
and 
.This post has been edited 190 times. Last edited by Virgil Nicula, Jan 7, 2016, 12:36 PM
August 2018
January 2018