Let's make a triangle with the solution

Algebra Level 5

Let $T_1$ be a triangle with angles $57^{\circ}, 60^{\circ}, 63^{\circ}$ and sides $a, b,c$ . Let $x,y,z$ be positive numbers such that: $\begin{aligned} x(y+z-x)=a^2, \\ y(z+x-y)=b^2, \\ z(x+y-z)=c^2. \end{aligned}$ It is known that $x,y,z$ represent the sides of a triangle $T_2$ . Find the measure (in degrees) of the greatest angle of $T_2$ .

The answer is 66.

2 solutions

Mark Hennings
Aug 27, 2013

Suppose that the angles of $T_1$ are $A,B,C$ . Note that $\begin{array}{rcl} 2ab\cos C & = & a^2 + b^2 - c^2 \; = \; 2xy - x^2 - y^2 + z^2 \\ & = & z^2 - (x - y)^2 \\ a^2b^2 & = & xy\big(z^2 - (x-y)^2\big) \; = \; 2abxy\cos C \\ ab & = & 2xy\cos C \\ ab\sin C & = & xy\sin2C \end{array}$ Similarly, $ac\sin B = xz\sin 2B$ and $bc\sin A = yz\sin2A$ . If $\Delta_1$ is the area of $T_1$ , then $yz\sin2A \; = \; xz\sin2B \; = \; xy\sin2C \; = \; 2\Delta_1$ and hence $\frac{\sin2A}{x} \; = \; \frac{\sin2B}{y} \; = \; \frac{\sin2C}{z}$ Since $x,y,z$ are the sides of a triangle $T_2$ and $A,B,C$ are all acute, the Sine Rule now tells us that the angles of $T_2$ are $180^\circ-2A$ , $180^\circ-2B$ , $180^\circ-2C$ . Note that the area $\Delta_2$ of $T_2$ is equal to the area $\Delta_1$ of $T_1$ .

In our case, the largest angle in $T_2$ is $(180-2\times57)^\circ = 66^\circ$ .

If we did not want just to say that the angles of $T_2$ must be these because they fit the pattern of the Sine Rule, we could push things through fully. We know that $x \; = \; R\sin2A \qquad y \; = \; R\sin2B \qquad z \; = \; R\sin2C$ for some $R>0$ . Now $\begin{array}{rcl} \sin^2x &+& \sin^2y \;-\; \sin^2(x+y)\\ & = & \sin^2x + \sin^2y - (\sin x \cos y + \cos x \sin y)^2 \\ & = & \sin^2x(1-\cos^2y) - 2\sin x \cos x \sin y \cos y\\ & & {} + (1 - \cos^2x)\sin^2y \\ & = & 2\sin^2x\sin^2y - 2\sin x \cos x \sin y \cos y \\ & = & 2\sin x \sin y(\sin x \sin y - \cos x \cos y) \\ & = & -2\sin x \sin y\cos(x+y) \end{array}$ and hence $\begin{array}{rcl} \sin^22B + \sin^22C - \sin^22A & = & -2\sin2B\sin2C\cos(2B+2C) \\ & = & -2\sin2B\sin2C\cos(360^\circ-2A) \\ & = & -2\sin2B\sin2C\cos2A \end{array}$ Thus, if $X,Y,Z$ are the angles of $T_2$ , then $\begin{array}{rcl}\cos X & = & \frac{y^2+z^2-x^2}{2yz} \; = \; \frac{R^2(\sin^22B + \sin^22C - \sin^22A)}{2R^2\sin 2B\sin 2C} \\ & = & -\cos2A \; = \;\cos(180^\circ-2A) \end{array}$ so that $X = 180^\circ - 2A$ and, similarly, $Y=180^\circ-2B$ and $Z = 180^\circ-2C$ .

Moderator note:

Great job! You really got to the bottom of this one!

We can even do this at least semi-geometrically (and at the same time prove the existence of triangle $T_2$ , rather than just accept its existence).

Let angles $A,B,C$ all be acute. Let $A',B',C'$ be the feet of the perpendiculars from $A,B,C$ to $BC,AC,AB$ respectively. Since $A,B,C$ are all acute, $A'$ lies within $BC$ , $B'$ lies within $AC$ and $C'$ lies within $AB$ . It is fairly standard that all four triangles $ABC$ , $AB'C'$ , $A'BC'$ and $A'B'C$ are similar. Thus the triangle $A'B'C'$ has angles $A' = 180^\circ-2A \qquad B' = 180^\circ - 2B \qquad C' = 180^\circ - 2C$ and sides $a' = a\cos A \qquad b' = b\cos B \qquad c' = c\cos C$ Then $\begin{array}{rcl} b'+c'-a' & = & b\cos B + c\cos C - a\cos A \\ & = & \frac{b(a^2+c^2-b^2)}{2ac} + \frac{c(a^2+b^2-c^2)}{2ab} - \frac{a(b^2+c^2-a^2)}{2bc} \\ & = & \frac{b^2(a^2+c^2-b^2) + c^2(a^2+b^2-c^2) - a^2(b^2+c^2-a^2)}{2abc} \\ & = & \frac{(a^2+b^2-c^2)(a^2+c^2-b^2)}{2abc} \\ & = & 2a\cos B \cos C \\ a'(b'+c'-a') & = & 2a^2\cos A \cos B \cos C \end{array}$ and, similarly, $b'(c'+a'-b') = 2b^2\cos A \cos B \cos C$ and $c'(a'+b'-c') = 2c^2\cos A \cos B \cos C$ .

Thus, since $A,B,C$ are all acute, we can scale the triangle $A'B'C'$ , forming a triangle $T_2$ with sides $x = \frac{a'}{\sqrt{2\cos A \cos B \cos C}} \;,\;y = \frac{b'}{\sqrt{2 \cos A \cos B \cos C}} \;,\; z = \frac{c'}{\sqrt{2\cos A \cos B \cos C}}$ and we obtain the correct identities for $x,y,z$ , while retaining the angles $180^\circ-2A,180^\circ-2B,180^\circ-2C$ .

Hmmh. Is there a geometric construction which obtains $T_2$ directly, instead of this geometric construction of the orthic triangle $A'B'C'$ and then rescaling?

Mark Hennings - 7 years, 9 months ago

The triangle $T_2$ can be constructed, since $\frac{1}{2\cos A \cos B \cos C} \; = \; \frac{R}{r'}$ where $R$ is the outradius of the original triangle $T_1$ , and $r'$ is the inradius of the orthic triangle $A'B'C'$ . Thus we need to construct $T_2$ , a triangle similar to the orthic triangle with inradius $\sqrt{Rr'}$ . This can be done:

alt text

Mark Hennings - 7 years, 9 months ago

Very thorough solution. Good job, Mark H.

Michael Tong - 7 years, 9 months ago

Oh my god! Magnifico! I say make that triangle your profile pic, Mark H. :)

A Brilliant Member - 7 years, 9 months ago

Alex Wice
Aug 27, 2013

In any triangle, by the Cosine Law we know:

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Now square both sides and substitute all occurrences of $a^2$ , $b^2$ , and $c^2$ with the given equalities. After reducing, you get

$\cos^2 C = \frac{(z^2 - x^2 - y^2 + 2xy)}{4xy}$ and so $1 - 2\cos^2 C = \frac{x^2 + y^2 - z^2}{2xy} = \cos Z$

where the last equality follows from the cosine law on the triangle with sidelengths $x,y,z$ .

Finishing, we have $\cos Z = 1 - 2\cos^2 C$ , and $\cos Z = -\cos 2C = \cos(180-2C)$ from the double angle and the formula $\cos (-x) = -\cos x$ .

So indeed $\cos X, \cos Y, \cos Z = \cos 54^\circ, \cos 60^\circ, \cos 66^\circ$ , in some order. The result follows. $\fbox{ 66 }$

$\cos(-x)=-\cos x$

I always thought it was:

$\cos (-x) = \cos x$ .

Did you want to write:

$\cos (180^\circ -x)=-\cos x$ ?

Mursalin Habib - 7 years, 9 months ago

I believe that was a typo. Note that the author told $- \cos \ 2C= \cos \ (\pi -2C)$ ,which seems to be an application of the identity $\cos \ \theta + \cos \ (\pi - \theta)= 0$ .

Sreejato Bhattacharya - 7 years, 9 months ago

Let's make a triangle with the solution

The answer is 66.

2 solutions

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