Prove that $\angle ABC <$ $ 84^\circ$.

Question

Angles A, B and C meet the circle at P, Q and R respectively.

I made the picture below, but what would the solution be?

Answer 1

Let $a,b,c$ be the sides of the given triangle $\Delta =\Delta ABC$, we denote by $p=(a+b+c)/2$ its half-perimeter, by $R,r$ the radius of the circumcircle, respectively of the inscribed circle, and by $S$ the area of $\Delta$.

Euler's relation, wiki link, gives $$r^2=R(R-2r)\ .$$ The corresponding equation of second degree is $0=R^2-2rR-r^2$, giving $R/r=1\pm \sqrt 2$, and we of course take only the positive choice. It is time to norm, let us consider below the case with $$R = 1\ ,\qquad r=\sqrt 2-1\ .$$

We show first algebraically that under these circumstances there is exactly one solution of the given problem (with $b>c$). Then we explore also a geometric path.

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From $\frac a{\sin A}=2R$ we obtain $a=2R\sin 60^\circ=\sqrt 3$.

Then we also have by the generalized theorem of Pythagoras a relation between $b,c$: $$ b^2+c^2-bc = b^2 +c^2-2bc\cos A=a^2 =3\ .$$ We need one more equation. The area $S$ appears in the formulas involving $r,R$ $\displaystyle R=\frac {abc}{4S}$, and $\displaystyle S = pr$, joining them, we get $$ 2(a+b+c)(\sqrt 2-1) = 4p\; rR = 4RS = abc\ . $$ This is the second equation in $b,c$. The two homogenous equations can be written simpler using the sum $\Sigma =b+c$, and the product $\Pi=bc$ of the two unknowns $b,c$, we get the system $$ \left\{ \begin{aligned} \Sigma^2-3\Pi &= 3\ ,\\ 2(\sqrt 3+\Sigma )(\sqrt 2-1) &= \sqrt 3\Pi\ . \end{aligned} \right. $$ We substitute $\Pi$ from the second equation in the first one, eliminate it, and obtain for $\Sigma$ the equation: $$ \Sigma^2 - 2\sqrt 3(\sqrt 3+\Sigma )(\sqrt 2-1) = 3 \ .$$ The roots are $\Sigma = \sqrt3(\sqrt 2-1)\pm\sqrt6$. Only the positive solution counts, $\Sigma=\sqrt3(2\sqrt 2-1)$. Then $\Pi = 8-4\sqrt 2$. Then $\Sigma^2-4\Pi = 4\sqrt 2-5= (\sqrt 2-1)^2(2\sqrt 2+1)$. We get the explicit formulas for $b,c$, $$ b,c =\frac 12\Big(\ \sqrt3(2\sqrt 2-1)\pm (\sqrt 2-1)\sqrt{2\sqrt 2+1}\ \Big)\ . $$

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Please allow here a quick computer check, so that i can go on.

sage: R, r, a, Sigma, Pi = 1, sqrt(2)-1, sqrt(3), sqrt(3)*(2*sqrt(2) - 1), 8-4*sqrt(2)
sage: b = ( sqrt(3)*(2*sqrt(2)-1) + (sqrt(2)-1)*sqrt(2*sqrt(2)+1) ) / 2
sage: c = ( sqrt(3)*(2*sqrt(2)-1) - (sqrt(2)-1)*sqrt(2*sqrt(2)+1) ) / 2
sage: ( b^2 + c^2 - b*c ).n()
3.00000000000000
sage: ( b+c - Sigma ).n()
0.000000000000000
sage: ( b*c - Pi ).n()
8.88178419700125e-16
sage: ( 2*(a+Sigma)*r*R - a*Pi ).n()
1.77635683940025e-15
sage: ( arcsin( b/2*R ) / pi * 180 ).n()
83.9057117814039

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Well, numerically the angle in $B$ in our triangle is slightly bigger than $83.9^\circ$, and we have to show that it is $<84^\circ$. This is a very close approximation.

Can we really find a (simple) geometric argument (instead of the above algebraic computation)?

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Some geometric thoughts to construct the triangle are as follows.

We draw a circle of radius one, $R=1$, centered in a point $O$, consider the points $M,B,C$ on it so that $\Delta MBC$ is equilateral, and search the "right place" on the same circle for the vertex $A$, so that the circle inscribed in $\Delta ABC$ passes through $O$. We assume the knowledge of its existence and uniqueness.

Let $L$ be the symmetric of $M$ w.r.t. $O$, so $ML$ is a diameter. Then $I$ is on $AL$, since $L$ is on the angle bisector $AI$ of the angle $A$, since it divides the arc $\overset \frown{BC}$ in two equal arcs. So $I$ is on the circle centered in $O$ with radius $r=\sqrt 2-1$.

Let $D,E,F$ be the projections of $I$ on the sides $BC$, $CA$, $AB$. Then $\Delta AIE=\Delta AIF$ have the angles $30^\circ$, $60^\circ$, $90^\circ$, which implies $AI=2IE=2IF=2r$.

In the triangle $\Delta BIL$ the angles in $B$ and $I$ are equal, so $IL=BL=BO=R=1$.

So $LA=1+2r=1+2(\sqrt 2-1)=2\sqrt 2-1< 2$. We can now construct $A$ in this way. We can compute now explicitly also $AM$, well $AM^2=4\sqrt 2-5$.

The fact that $\hat B=\widehat{ABC}=\widehat{ABM}+\widehat{MBC} =\widehat{ABM}+60^\circ<84^\circ$ is equivalent to $$ \widehat{ABM}=\widehat{ALM}<24^\circ\ ,$$ and we may work to get a proof by comparing $AM$ with the side $XM$ of the corresponding regular polygon with $\widehat{XLM}=24^\circ$.

(Or we may compute the chord corresponding to the triple of the above angle, and use the knowledge of the side of the $5$-gon to compare explicitly.)

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I may come back with a picture and more details if this is still relevant, please mention the point to put accent on in a short comment.

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Below there is a geogebra picture of the construction of $\Delta ABC$ following the above description.

Since the figure is overloaded, here is the order of introducing the objects.

• We draw a circle with (blue) radius $R=1$ centered in a point $O$, and draw a point $M$ on it. We call this circle $(O)$ below.
• We construct the regular polygons with three and four vertices, inscribed in the circle $(O)$ with one vertex in $M$. The equilateral triangle constructed is denoted by $\Delta MBC$. From the square we need later only the construction of $\sqrt 2$. Let $L$ be so that $ML$ is a diameter, so $ML=2R=2$.
• We construct as in the figure the length $2\sqrt 2-1< 2$ and draw a big circle centered in $L$ with this radius. It intersects the circle $(O)$ in two points, we denote by $A$ the intersection point closer to $B$.
• We construct $I$, the incenter of $\Delta ABC$.
• Then $O$ is on the circle centered in $I$ and radius $r=\sqrt 2-1$.

Geogebra can mark angles with their value, so asking it for this service, we get the approximative value shown in the figure, $\color{magenta}{\approx 83.9057^\circ}$.

Answer 2

The figure shows $\triangle ABC$ with incenter $I$, inradius $r$, circumcenter $K$, and circum-half-radius $s:=|KS|$, where $S$ is the midpoint of $\overline{BC}$. (Since $\angle BKC=2\angle BAC=120^\circ$, we conclude $\triangle CKS$ is $30^\circ$-$60^\circ$-$90^\circ$, so that $|KS|$ is half the circumradius $|KC|$.)

Now, the amount that $\angle B$ falls short of $90^\circ$ is the same the amount that $\angle C$ exceeds $30^\circ$; that's $\theta := \angle KCA$, and we have $$\cos\theta = \frac{|AC|/2}{|KC|}=\frac{|AT|+|CT|}{4s}=\frac{|AT|+|CS|+|SR|}{4s}=\frac{(r+s)\sqrt{3}+p}{4s} \tag{1}$$ where $$p^2 = r^2-(s-r)^2=s(2r-s) \tag{2}$$

From here, it's just a bit of algebra.

As in @dan_fulea's answer, we invoke Euler's formula, with circumradius $2s$ and incenter-to-circumcenter distance $r$:

$$r^2 = 4s(s-r) \quad\to\quad r = 2s (\sqrt{2}-1) \quad\to\quad p = s \sqrt{4 \sqrt{2}-5} \tag{3}$$

(where I've discarded negative $r$ and $p$ candidates) so that

$$\cos\theta = \frac14\left(\;(2\sqrt{2}-1)\sqrt{3}+\sqrt{4\sqrt{2}-5}\;\right) \tag{4}$$

It happens that $\theta = 6.09429\ldots^\circ$, from which the desired result follows. However, demonstrating this with a clean geometric argument eludes me. (See @dan's discussion.) Given that $\cos 6^\circ-\cos\theta \approx 0.00017$, there isn't a great deal of room to play.

Answer 3

$\begin{array}{} \text{Euler's relation} & IO=\sqrt{R(R-2r)} & IO=r=1 & R=1+\sqrt{2} \end{array}$

$\begin{array}{} sin(\frac{ π }{6})=\frac{1}{2} & AI=2 & AI^2=1+y_{A}^2 & A=(1,y_{A}=\sqrt{3}) \end{array}$

$\begin{array}{} {O}=λ ∩ ω & O=(h,k) \end{array}$

$\left\{ \begin{array}{} x^2+y^2=1 \\ (x-1)^2+(y-\sqrt{3})^2=(1+\sqrt{2})^2 \end{array} \right.$

$\begin{array}{} h=\frac{1-\sqrt{2}+\sqrt{3+6\sqrt{2}}}{4} & k=\frac{-3+3\sqrt{2}+\sqrt{3+\sqrt{2}}}{4\sqrt{3}} \end{array}$

$\begin{array}{} {C}=ψ∩t & \left\{ \begin{array}{} x=1 \\ (x-h)^2+(y-k)^2=(1+\sqrt{2})^2 \end{array} \right. \end{array}$

$\begin{array}{} y_{C}=k+\sqrt{2h-h^2+2\sqrt{2}+2} & y_{C}=\frac{3\sqrt{3}+3\sqrt{6}+\sqrt{9+18\sqrt{2}}}{6} \end{array}$

$\begin{array}{} AC=\sqrt{3}+y_{C} & AC=\frac{3\sqrt{3}+\sqrt{6}+\sqrt{1+2\sqrt{2}}}{2} \end{array}$

$\begin{array}{} ϕ= \angle(AOB)& θ=\frac{ϕ}{2} & AO=OC=1+\sqrt{2} \end{array}$

$\begin{array}{} AC^2=AO^2+OC^2-2·AO·OC·cos(ϕ) & cos(ϕ)=1-\left( \frac{AC}{OC} \right) ^2 \end{array}$

$cos(ϕ)=\frac{-7+4\sqrt{2}-5\sqrt{3+6\sqrt{2}}+3\sqrt{6+12\sqrt{2}}}{4}$

$\begin{array}{} θ=\angle(ABC) & θ=\frac{ϕ}{2} \end{array}$

$θ=\frac{1}{2}arccos\left( \frac{-7}{4}+\sqrt{2}-\frac{5}{4}\sqrt{3+6\sqrt{2}}+\frac{3}{4}\sqrt{6-12\sqrt{2}} \right)·\frac{180}{π}=83.905711$

$\begin{array}{} \text{checking} & θ<\frac{7π}{15} & ϕ=2θ<\frac{14π}{15} \end{array}$

$\begin{array}{} cos(ϕ)>cos(\frac{14π}{15}) & cos(ϕ)>cos(\frac{π}{15}) & cos(\frac{π}{15}>-cos(ϕ)) \end{array}$

$\frac{1}{8}\left( -1+\sqrt{5}+\sqrt{6(5+\sqrt{5}} \right) >\frac{-1}{4}\left( -7+4\sqrt{2}-5\sqrt{3+6\sqrt{2}}+3\sqrt{6-12\sqrt{2}} \right)$

how to prove it exactly?

$\begin{array}{} \text{True (numerical verification)} & \text{difference} & 6.896×10^{-4} \end{array}$

($λ$) Incircle of a $triangleABC$, (locus of point $O$) with center $I=(0,0)$ and radius $r=1$, $T=(1,0)$, ($t$) vertical line $x=1$ (locus of points $A$ and $C$), [$\frac{π}{3}$ angle construction] ($μ$) circle with center $I$ and radius=2, $A=(1,\sqrt{3})$ intersection $μ$ and $t$, $AI=2$, $D=(-2,0)$ intersection $μ$ and $x-axis$, $AD$, $\angle(IAT)=\frac{π}{6}$, $sin(\frac{π}{6})=\frac{1}{2}$, $\angle(DAI)=\frac{π}{6}$, ($ω$) circle with center $A$ and $R=1+\sqrt{2}$, (Euler's relation), $O=(h,k)$ intersection $λ$ and $ω$, ($ψ$) circle by $A$ with center in $O$ (locus points $A$, $B$ and $C$), $B$ intersection $AD$ and $ψ$, $C$ intersection $ψ$ and $t$, $BC$