find the range of angle BAC

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find the range of angle BAC

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Source: China TST 1989, problem 6; China North MO

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3647 posts

orl

#1 h Jun 27, 2005, 9:43 AM • 2 Y

Y by Adventure10, Mango247

$AD$ is the altitude on side $BC$ of triangle $ABC$ . If $BC+AD-AB-AC = 0$ , find the range of $\angle BAC$ .

Alternative formulation. Let $AD$ be the altitude of triangle $ABC$ to the side $BC$ . If $BC+AD=AB+AC$ , then find the range of $\angle{A}$ .

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#2 h Jun 30, 2005, 12:34 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

well, sin(A) = sin(B) + sin(C) - sin(B)sin(C), so 0< A < 180 is all fine, right?

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#3 h Jun 30, 2005, 4:18 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

I write your relation such: ( 1 - sinB ) ( 1 - sinC ) = ( 1 - sinA ) and is it easy a.s,o ???

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6555 posts

#4 h Jun 30, 2005, 7:55 AM • 1 Y

Y by Adventure10

My solution uses another approach:

orl wrote:

$AD$ is the altitude on side $BC$ of triangle $ABC$ . If $BC + AD - AB - AC = 0$ , find the range of $\angle BAC$ .

First we show:

Lemma 1. Let D be the foot of the altitude to the side BC of a triangle ABC. Then, BC + AD - AB - AC = 0 holds if and only if $\tan\frac{B}{2}+\tan\frac{C}{2}=1$ .

Proof of Lemma 1. Let $s=\frac{a+b+c}{2}$ be the semiperimeter, $\Delta$ the area and $r_a$ the A-exradius of triangle ABC. Then, on one hand, $\Delta=\frac12\cdot a\cdot AD$ , since AD is the altitude to the side BC of triangle ABC, but on the other hand, $\Delta=\left(s-a\right)r_a$ by a well-known formula. Hence, $\frac12\cdot a\cdot AD=\left(s-a\right)r_a$ ; equivalently, $AD=\frac{2\left(s-a\right)r_a}{a}$ . On the other hand, b + c - a = (a + b + c) - 2a = 2s - 2a = 2 (s - a). Thus,

$BC+AD-AB-AC=BC+\frac{2\left(s-a\right)r_a}{a}-AB-AC=a+\frac{2\left(s-a\right)r_a}{a}-c-b$
$=\frac{2\left(s-a\right)r_a}{a}-\left(b+c-a\right)=\frac{2\left(s-a\right)r_a}{a}-2\left(s-a\right)$
$=2\left(s-a\right)\frac{r_a-a}{a}$ .

Since s - a > 0, we thus have BC + AD - AB - AC = 0 if and only if $r_a-a=0$ , what is equivalent to $a=r_a$ .

Now, let the A-excircle of triangle ABC have the center $I_a$ and touch the side BC at a point X. Then, $I_aX=r_a$ , and $I_aX\perp BC$ . But, being the A-excenter of triangle ABC, the point $I_a$ lies on the external angle bisector of the angle ABC, and thus $\measuredangle I_aBX=90^{\circ}-\frac{B}{2}$ . Hence, in the right-angled triangle $BXI_a$ , we have $BX=I_aX\cdot\cot\measuredangle I_aBX=r_a\cdot\cot\left(90^{\circ}-\frac{B}{2}\right)=r_a\cdot\tan\frac{B}{2}$ . Similarly, $CX=r_a\cdot\tan\frac{C}{2}$ . Thus, $a=BC=BX+CX=r_a\cdot\tan\frac{B}{2}+r_a\cdot\tan\frac{C}{2}=r_a\cdot\left(\tan\frac{B}{2}+\tan\frac{C}{2}\right)$ . Hence, $a=r_a$ is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$ . But we saw before that BC + AD - AB - AC = 0 is equivalent to $a=r_a$ . Hence, BC + AD - AB - AC = 0 is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$ , and Lemma 1 is proven.

Another lemma now:

Lemma 2. For a given angle x such that 0° < x < 90°, the existance of two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$ is equivalent to $45^{\circ}<x\leq 2\arctan\frac12$ .

Note. $2\arctan\frac12\approx 53,13^{\circ}$ .

Proof of Lemma 2. First, we show that if there exist two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$ , then $45^{\circ}<x\leq 2\arctan\frac12$ .

In fact, the function f(t) = tan t is convex on the interval [0°; 90°[; hence, $\tan y+\tan z\geq 2\tan\frac{y+z}{2}$ . In other words, $1\geq 2\tan\frac{x}{2}$ . Hence, $\frac12\geq\tan\frac{x}{2}$ , so that $\arctan\frac12\geq\frac{x}{2}$ and $2\arctan\frac12\geq x$ . On the other hand, by the tangens addition formula, $\tan\left(y+z\right)=\frac{\tan y+\tan z}{1-\tan y\tan z}$ . In other words, $\tan x=\frac{1}{1-\tan y\tan z}$ . Since tan y and tan z are positive (as 0° < y < 90° and 0° < z < 90°), we have 1 - tan y tan z < 1; but tan x is also positive (since 0° < x < 90°), and thus, tan x > 1, and thus x > 45°. Combining this with $2\arctan\frac12\geq x$ , we obtain $45^{\circ}<x\leq 2\arctan\frac12$ .

Remains to prove the converse: If $45^{\circ}<x\leq 2\arctan\frac12$ , then there exist two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$ .

In fact, first define the angles y and z as follows: y = 0°, z = x. This will be called the "starting position". Of course, y + z = x is satisfied in this position, but $\tan y+\tan z=1$ does not hold.

Now, let's start continuously increasing y and decreasing z at the same speed, so that the sum y + z = x remains invariant. Then, the angles y and z come closer and closer to each other, and at the end both of them come together: $y=z=\frac{x}{2}$ . This will be called the "ending position".

Now look at what happens with the value of tan y + tan z while we increase y and decrease z. Since the function f(t) = tan t is continous on the interval [0°; 90°[, this value of tan y + tan z behaves continuously. In the "starting position", the value of tan y + tan z equals tan 0° + tan x = tan x; this is > 1, since x > 45°. In the "ending position", the value of tan y + tan z equals

$\tan\frac{x}{2}+\tan\frac{x}{2}=2\tan\frac{x}{2}\leq2\tan\arctan\frac12$ (since $x\leq 2\arctan\frac12$ yields $\frac{x}{2}\leq\arctan\frac12$ )
$=2\cdot\frac12=1$ .

Hence, by the intermediate value theorem, somewhere between the "starting position" and the "ending position", the expression tan y + tan z must take the value 1. Thus, we have proved the existence of two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$ . Hence, Lemma 2 is proven.

Now, the problem becomes immediate: After Lemma 1, the condition BC + AD - AB - AC = 0 is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$ . Hence, we are looking for the range of angles A such that 0° < A < 180° and such that there exist angles B and C with 0° < B < 180°, 0° < C < 180° and A + B + C = 180° (these are the conditions for the existence of a triangle with angles A, B, C) which satisfy $\tan\frac{B}{2}+\tan\frac{C}{2}=1$ . Set $x=90^{\circ}-\frac{A}{2}$ , $y=\frac{B}{2}$ , $z=\frac{C}{2}$ ; then, the conditions 0° < A < 180°, 0° < B < 180°, 0° < C < 180° rewrite as 0° < x < 90°, 0° < y < 90°, 0° < z < 90°, the condition A + B + C = 180° rewrites as y + z = x (since $y+z-x=\frac{B}{2}+\frac{C}{2}-\left(90^{\circ}-\frac{A}{2}\right)=\frac{A+B+C}{2}-90^{\circ}=\frac{A+B+C-180^{\circ}}{2}$ ), and the condition $\tan\frac{B}{2}+\tan\frac{C}{2}=1$ rewrites as $\tan y+\tan z=1$ . Now, by Lemma 2, the existence of two angles y and z satisfying these conditions is equivalent to $45^{\circ}<x\leq 2\arctan\frac12$ . Since $x=90^{\circ}-\frac{A}{2}$ , this rewrites as $45^{\circ}<90^{\circ}-\frac{A}{2}\leq 2\arctan\frac12$ . Upon subtraction from 90°, this becomes $45^{\circ}>\frac{A}{2}\geq 90^{\circ}-2\arctan\frac12$ , and multiplication with 2 transforms this into $90^{\circ}>A\geq 2\left(90^{\circ}-2\arctan\frac12\right)$ . This rewrites as $90^{\circ}>A\geq 180^{\circ}-4\arctan\frac12$ . Thus, the range of our angle A is $\left[180^{\circ}-4\arctan\frac12;\;90^{\circ}\right[$ .

Note that $180^{\circ}-4\arctan\frac12\approx 73,74^{\circ}$ .

Darij

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frejer

#5 h Jan 26, 2006, 3:20 AM • 1 Y

Y by Adventure10

I think you thought too complex( although you solution is not very special)
My solution is bases on the problem:
find all A in (0,pi) that the equation f(x)=sinx+sin(x+A)-sinx sin(x+A)- sinA has solution x in (0,pi-A).
simply by using derivative.

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#6 h Aug 15, 2006, 4:54 PM • 2 Y

Y by Adventure10, Mango247

If $BC = a$ , $CA = b$ and $AB = c$ then either $a \le \min\{b,c\}$ or $a \ge \max\{b,c\}$ . Does this lead anywhere?

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