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I am trying to solve this geometry problem from an exam. The exam is supposed to be 3 hours long, and this is supposed to be 1 out of 10 problems. So given that, the solution should be something quick. However, the only solution I've been able to find is extremely convoluted and realistically I'd never be able to give an answer in less than 3 hours. So, I'm looking for a simpler solution ( the intended solution ) which I should have been able to do in the exam. Here's the problem:

Consider this illustration:
image 1

In the image, the outer shape is a square with side length $8$, a circle is drawn so that is is tangent to two sides of the square at points $G$ and $F$, then the line $AH$ is drawn so that it is tangent to the circle at point $I$ and also such that the length of the segment $AI$ is also equal to $8$. Lastly the line $BJ$ is drawn such that it is tangent to the circle at point $K$. Find the area of the blue triangle, that is the area of $\triangle{AJB}$.

My approach so far is to divide the square into 9 different areas : image 2
I also added the segment $EL$ such that it is the continuation of the line $GE$. From here on out when I say $a_{i}$ , I am referring to the area of region $a_{i}$ , $r$ will be the radius of the circle, $x$ will be the length of segment $JI$ and $y$ will be the length of segment $IH$.
So we want to find $a_{9}$

Notice then that:
$$ a_{1} = a_{2} = \frac{r(8-r)}{2} \\ a_{3} = r^2 \\ a_{4} = a_{5} = \frac{ry}{2} \\ a_{6} = a_{7} = \frac{rx}{2} \\ a_{8} = \frac{8(8-r-y)}{2} = 4(8-r-y) \\ a_{1} + \cdots + a_{9} = 8*8 = 64 \\ a_{1} + \cdots + a_{8} = 4r + ry + rx + 32 - 4y = 32 + r(4 + x) + y(r - 4) \\ a_{9} = 64 - (32 + r(4 + x) + y(r - 4)) = 32 - r(4 + x) + y(4 - r) $$
So from here really all we need to do is find $x, y, r$ and problem solved, right? Finding $r$ is pretty easy, notice that line segment $AE$ can be thought of as the hypotenuse of $\triangle{AIE}$ and the hypotenuse of $\triangle{AEL}$. Using Pythagoras, we have $8^2 + r^2 = (8-r)^2 + (8-r)^2$. Solving for $r$ we get two solutions: $16+8\sqrt{3}, 16-8\sqrt{3}$ and since $r$ is lesser than $8$, then the second one is our value of $r$.

To find the value of $y$, notice that by Pythagoras on $\triangle{ACH}$, we have: $(8 + y)^2 = 8^2 + (8-r-y)^2$, and since we know the value of $r$ we can solve for $y$ and we get $y = \frac{16\sqrt{3}-24}{3}$.

From here, I don't actually know how to find the value of $x$, so I went back to our expression for the desired area $a_{9} = 32 - r(4 + x) + y(4 - r)$, plugged in the values of $r$ and $y$ to make the area a function of $x$, then I found another function of $x$ for the area, using Heron's formula. Notice the perimeter of the desired triangle is equal to $8 + (8 - x) + (8 - r + x)$, and when we plugged for $r$ we get that the semiperimeter $S$ should be equal to $S = 4 \sqrt{3} + 4$. Since the sides of the triangle have length $8, 8-x, 8-r+x$ and since we have precise values for the semiperimeter and $r$, we can sort of make a function for the area based on values of $x$. If we make these two functions equal to each other and solve the equation, we get two values which are possible candidates for $x$.
Problem is, these values are super convoluted, involve really large numbers and a bunch of radical expressions, which is why I gave up on this solution, realistically if it wasn't for Symbolab solving the equation for me, I probably wouldn't have time to do so in the exam.

So that's why I am writing this post : I would like for you guys to help me come up with a concrete value for $x$ in an easier way, or maybe a completely different way of computing the area that doesn't involve $x$, anything that is simpler than what I've been doing.

Thank you so much for reading all of this, I know that it's kind of a long post.

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    $\begingroup$ +1 for the description of the picture $\endgroup$ Commented Jan 23 at 3:00
  • $\begingroup$ Since $AI = 8$, you have an equilateral triangle there and so $\angle IAB = 60^\circ$ and if you let $AI$ intersect $BD$ at some point $O$, then you get a $30-60-90$ triangle . I think there is probably a solution in there somewhere like that. $\endgroup$
    – dezdichado
    Commented Jan 23 at 5:03
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    $\begingroup$ @dezdichado I can see why $\triangle {AIB}$ would be isosceles, but I don't see why it should be equilateral, could you give more details? $\endgroup$
    – zlaaemi
    Commented Jan 23 at 5:17
  • $\begingroup$ Are you aware that the angle $ABK$ is $45°$? It must be, because the line $AKED$ is the diagonal of a perfect square. $\endgroup$
    – Dominique
    Commented Jan 23 at 10:29
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    $\begingroup$ @Dominique: But in the drawing, $K$ is definitely not on the diagonal! $\endgroup$
    – TonyK
    Commented Jan 23 at 12:50

3 Answers 3

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In such issues, you can "deconvolute" the difficulties, first of all by using the power of point $A$ with respect to circle (C) in 2 different ways :

$$P(A,(C))=AE^2-r^2=PI^2 \tag{1}$$

Knowing that $AE=AD-ED=\sqrt{2}(8-r).$

(1) gives :

$$2(8-r)^2-r^2=64$$

This quadratic equation has the unique compatible solution

$$r=8(2-\sqrt{3})\tag{2}$$ (as you have obtained).


Edit : Here is the end of the computation.

enter image description here

Let us compute the tangent of the two angles at the base of triangle $ABJ$ :

$$\tan(\angle BAJ)=\tan(\alpha)=\tan(\frac{\pi}{4}+\arctan \frac{r}{8})=\frac{1+\frac{r}{8}}{1-\frac{r}{8}}=\frac{8+r}{8-r}\tag{1}$$

$$\tan(\angle JBA)=\tan(\frac{\pi}{2}-2\beta)=\cot(2 \beta)=\frac{1}{\tan(2 \beta)}=\frac{1- (\tan \beta)^2}{2 \tan \beta}=\frac{8(4-r)}{r(8-r)},\tag{3}$$

the last equality being a consequence of relationship :

$$\tan(\beta)=\frac{r}{8-r}\tag{3'}$$

If we denote by $P$ the foot of the altitude issued from $J$, and $h=JP$, we can write :

$$\begin{cases}AP&=&\frac{h}{\tan(\angle BJA)}\\ PB&=&\frac{h}{\tan(\angle JBA)}\end{cases}$$

adding them, we get, by dividing by $h$ :

$$\frac{8}{h}=\frac{r(8-r)}{8(4-r)}+\frac{8-r}{8+r}$$

from which (using (2))

$$h=\frac{8}{11}(6+\sqrt{3})\approx 5.623309678$$

is obtained and from there the area :

$$\frac{32}{11}(6+\sqrt{3})$$

of triangle $JAB$.

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  • $\begingroup$ I see how you can use the power of a point to compute $r$, but I am not quite sure how you would do it for $J$ and computing $x$, I tried it but I get equations that cancel out (get $0 = 0$), would you mind giving a bit more detail? $\endgroup$
    – zlaaemi
    Commented Jan 23 at 18:01
  • $\begingroup$ In fact (see my Edit), I have found a different way to obtain the result. $\endgroup$
    – Jean Marie
    Commented Jan 23 at 23:47
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    $\begingroup$ @zlaaemi FYI, using Jean's result, the area is $4h\approx 22.4932388$ while my answer's result is $a_9\approx 22.49323871$, i.e., they agree to $6$ decimal places. Since this answer's method is fairly different from mine, this not quite strongly indicates that we both have the correct answer, but also that your values of $y$ and $r$ are quite likely correct. $\endgroup$ Commented Jan 24 at 2:10
  • $\begingroup$ Thanks @Etemon for corrections. $\endgroup$
    – Jean Marie
    Commented Jan 24 at 8:14
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    $\begingroup$ @JohnOmielan I actually went through the trouble of computing a precise value of $h$ based on Jean's equation at the end, I got that $h = \frac{8(6+\sqrt{3})}{11}$, so $4h = \frac{32(6+\sqrt{3})}{11}$, which is exactly the same result you got, so I think it's safe to say that the methods are sound. $\endgroup$
    – zlaaemi
    Commented Jan 24 at 15:52
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You've made a great start. Continuing on, let

$$\alpha = \measuredangle HEG = \measuredangle HEI, \; \beta = \measuredangle BEF = \measuredangle BEK, \; \gamma = \measuredangle JEK = \measuredangle JEI$$

We then get

$$\tan(\alpha) = \frac{y}{r}, \;\; \tan(\beta) = \frac{8-r}{r}, \;\; \tan(\gamma) = \frac{x}{r}$$

where

$$\begin{equation}\begin{aligned} \frac{y}{r} & = \frac{\frac{16\sqrt{3}-24}{3}}{16 - 8\sqrt{3}} \\ & = \frac{2\sqrt{3} - 3}{3(2 - \sqrt{3})} \\ & = \frac{(2\sqrt{3} - 3)(2 + \sqrt{3})}{3(2 - \sqrt{3})(2 + \sqrt{3})} \\ & = \frac{4\sqrt{3} + 2(3) - 6 - 3\sqrt{3}}{3} \\ & = \frac{1}{\sqrt{3}} \end{aligned}\end{equation}$$

and

$$\begin{equation}\begin{aligned} \frac{8-r}{r} & = \frac{8 - (16 - 8\sqrt{3})}{16 - 8\sqrt{3}} \\ & = \frac{-8 + 8\sqrt{3}}{16 - 8\sqrt{3}} \\ & = \frac{-1 + \sqrt{3}}{2 - \sqrt{3}} \\ & = \frac{(-1 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} \\ & = \frac{-2 - \sqrt{3} + 2\sqrt{3} + 3}{4 - 3} \\ & = 1 + \sqrt{3} \end{aligned}\end{equation}$$

Using the formulae for the tangent of the sum and difference of angles in the Angle sum and difference identities section results in

$$\begin{equation}\begin{aligned} \tan(\alpha + \beta) & = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} \\ & = \frac{\frac{y}{r} + \frac{8-r}{r}}{1 - \left(\frac{y}{r}\right)\left(\frac{8-r}{r}\right)} \\ & = \frac{\frac{1}{\sqrt{3}} + (1 + \sqrt{3})}{1 - \left(\frac{1}{\sqrt{3}}\right)(1 + \sqrt{3})} \\ & = \frac{1 + \sqrt{3} + 3}{\sqrt{3} - (1 + \sqrt{3})} \\ & = -4 - \sqrt{3} \end{aligned}\end{equation}$$

Note that

$$2\alpha + 2\beta + 2\gamma = 360^{\circ} - 90^{\circ} \;\to\; \gamma = 135^{\circ} - (\alpha + \beta)$$

We then get

$$\begin{equation}\begin{aligned} \tan(\gamma) & = \tan(135^{\circ} - (\alpha + \beta)) \\ & = \frac{\tan(135^{\circ}) - \tan(\alpha + \beta)}{1 + \tan(135^{\circ})\tan(\alpha + \beta)} \\ & = \frac{-1 - (-4 - \sqrt{3})}{1 + (-1)(-4 - \sqrt{3})} \\ & = \frac{3 + \sqrt{3}}{5 + \sqrt{3}} \end{aligned}\end{equation}$$

Next, we have

$$\begin{equation}\begin{aligned} x & = r\tan(\gamma) \\ & = \frac{8(2 - \sqrt{3})(3 + \sqrt{3})}{5 + \sqrt{3}} \\ & = \frac{8(6 + 2\sqrt{3} - 3\sqrt{3} - 3)}{5 + \sqrt{3}} \\ & = \frac{8(3 - \sqrt{3})}{5 + \sqrt{3}} \\ & = \frac{8(3 - \sqrt{3})(5 - \sqrt{3})}{(5 + \sqrt{3})(5 - \sqrt{3})} \\ & = \frac{8(18 - 8\sqrt{3})}{25 - 3} \\ & = \frac{8(9 - 4\sqrt{3})}{11} \end{aligned}\end{equation}$$

Finally, using your formula for $a_9$, we end up with

$$\begin{equation}\begin{aligned} a_9 & = 32 - r(4 + x) + y(4 - r) \\ & = 32 - 8(2 - \sqrt{3})\left(4 + \frac{8(9 - 4\sqrt{3})}{11}\right) + \left(\frac{8(2\sqrt{3} - 3)}{3}\right)(4 - 8(2 - \sqrt{3})) \\ & = 8\left(4 - (2 - \sqrt{3})\left(\frac{116 - 32\sqrt{3}}{11}\right) + \left(\frac{2\sqrt{3} - 3}{3}\right)(-12 + 8\sqrt{3})\right) \\ & = \frac{8}{33}(132 - 3(2 - \sqrt{3})(116 - 32\sqrt{3}) + 4(11)(2\sqrt{3} - 3)(2\sqrt{3} - 3)) \\ & = \frac{8}{33}(132 - 3(232 - 64\sqrt{3} - 116\sqrt{3} + 96) + 44(12 - 12\sqrt{3} + 9)) \\ & = \frac{8}{33}(132 - 3(328 - 180\sqrt{3}) + 44(21 - 12\sqrt{3})) \\ & = \frac{8}{33}(72 + 12\sqrt{3}) \\ & = \frac{32(6 + \sqrt{3})}{11} \\ & \approx 22.49 \end{aligned}\end{equation}$$

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Let consider a coordinate system with origin in $E$ and axys parallel to the sides of the square. As you have found the radius of the circle is given by $$ r=8(2-\sqrt{3}) $$ and in this coordinate system the circle has equation $$ x^2+y^2=r^2 $$ The point $A$ has coordinates $A=(-8+r,-8+r)$ while $B=(r, -8+r)$

The equation for the line tangent to a circle with center in the origin at the point $(x_T,y_T)$ is $$ (y-y_T)y_T +(x-x_T)x_T=0 $$ (see e.g. this)

With this we can obtain the coordinates of the points $K=(x_K,y_K)$ and $I=(x_I,y_I)$ by solving $$ \left(-8+r-\sqrt{r^2-x_K^2}\right)\sqrt{r^2-x_K^2}+(r-x_K)x_K=0 $$ i.e. $$ \left(r^2+(8(1+\sqrt{3})^2\right)x_K^2-2r^3 x_K +\left(r^2-(8(1+\sqrt{3})^2\right)r^2=0 $$ By solving the quadratic equation and choosing the negative solution we obtain the equation for $x_K$. Plugging the value in $y_K=-\sqrt{r^2-x_K^2}$ you obtain the coordinates for $K$. Similarly you obtain the coordinates for $I$ (in this case there are less calculation as the resulting equation is linear in $x_I$)

With the coordinates for $K$ and $I$ you have the equation of the two tangent lines. This allow you to obtain the coordinates of their itersection i.e. $J$

$$ y_J=\frac{x_I}{x_Iy_K-x_K} \left(y_k^2-\frac{x_K}{x_I} y_I^2\right)-x_Iy_I $$ $$ x_J=-\frac{1}{x_I}(y_J-y_I)y_I-x_I $$ With the coordinate for $J$ you can now calculate the length of the two sides $AJ$ and $BJ$ and obtain the area of the triangle with Eron formula

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