Hard geometry problem: given a triangle with sides 16, 30, 34, find the area of the triangle introduced by the inscribed circle

Question

You are given a triangle $\triangle PQR$ with sides $16, 30, 34$. Let the incircle touch the sides of $\triangle PQR$ at $X,Y,$ and $Z$. Given that the ratio $[XYZ]/[PQR]$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, find $m+n$.

What I did first was to draw the diagram and I noticed that $PX=PZ, QX=QY, RY=RZ$ due to two intercepting lines both being tangent to a circle. Since triangles $PXZ, QXY, RYZ$ are isosceles I bisected them and got another right triangle after I tried to do similar triangles but got a little stuck.

My other friend used "barycentric coordinates", which I thought was completely unnecessary.

Any help would be appreciated.

Answer 1

You've made a good start, with the following showing how to finish (although with quite a bit work required) without using, at least explicitly, "barycentric coordinates" that your other friend used. First, the following is a diagram of the situation, with some additional lines, points and lengths included,

As shown, we have $|PQ|=16$, $|QR|=30$ and $|RP|=34$. Also, $O$ is the triangle's incenter. In addition, $A$, $B$ and $C$ are the intersection points of $OR$ and $ZY$, $OQ$ and $XY$, and $OP$ and $XZ$, respectively. Finally, $r$ is the radius of the incircle.

Since $PO$ is a common side and $|OZ| = |OX| = r$, then by the Pythagorean theorem, we have $|PX| = |PZ|$, as you've already stated. To get the outside triangle lengths shown, note that

$$\begin{equation}\begin{aligned} |PQ| + |QR| & = (|PX| + |XQ|) + (|QY| + |RY|) \\ 16 + 30 & = (|PZ| + |XQ|) + (|XQ| + |RZ|) \\ 46 & = (|PZ| + |RZ|) + 2|XQ| \\ 46 & = 34 + 2|XQ| \\ |XQ| & = 6 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Similarly, we get $|PX| = |PZ| = 10$ and $|ZR| = |YR| = 24$. Next, using your notation of square brackets meaning the area, using the area of a triangle is one half the base times the height, we get

$$\begin{equation}\begin{aligned} \text{[}PQR] & = [POQ] + [POR] + [QOR] \\ & = 8r + 15r + 17r \\ & = 40r \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Another way to get the area of a triangle is using Heron's formula, with $s$ (the semi-perimeter) here being $40$, with this giving

$$\begin{equation}\begin{aligned} \text{[}PQR] & = \sqrt{40(40-16)(40-30)(40-34)} \\ & = \sqrt{40(24)(10)(6)} \\ & = \sqrt{57\text{,}600} \\ & = 240 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Equating \eqref{eq2A} and \eqref{eq3A} shows $r = 6$. Thus, $|OR| = \sqrt{|RZ|^2 + 6^2}$. Next, note that $ZY$ is perpendicular to $OR$, and similarly with the other $2$ points. This is quite easy to prove, or you can just use that this is a property of kites. Thus, determining the area of $\triangle ORZ$ by using the half base times height in $2$ different ways, we have

$$\begin{equation}\begin{aligned} \frac{|RZ|\,|OZ|}{2} & = \frac{|OR|\,|AZ|}{2} \\ 6\,|RZ| & = \left(\sqrt{|RZ|^2 + 36}\right)|AZ| \\ |AZ| & = \frac{6\,|RZ|}{\sqrt{|RZ|^2 + 36}} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Using the Pythagorean theorem, we next get

$$\begin{equation}\begin{aligned} |OA| & = \sqrt{|OZ|^2 - |AZ|^2} \\ & = \sqrt{36 - \frac{36\,|RZ|^2}{|RZ|^2 + 36}} \\ & = \sqrt{\frac{36\,|RZ|^2 + 36^2 - 36\,|RZ|^2}{|RZ|^2 + 36}} \\ & = \frac{36}{\sqrt{|RZ|^2 + 36}} \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

Next, using \eqref{eq4A} and \eqref{eq5A},

$$\begin{equation}\begin{aligned} \text{[}OYZ] & = |AZ|\,|OA| \\ & = \left(\frac{6\,|RZ|}{\sqrt{|RZ|^2 + 36}}\right)\left(\frac{36}{\sqrt{|RZ|^2 + 36}}\right) \\ & = \frac{216\,|RZ|}{|RZ|^2 + 36} \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

Doing the same thing for the other $2$ triangles and adding their areas gives

$$\begin{equation}\begin{aligned} \text{[}XYZ] & = [OYZ] + [OYX] + [OXZ] \\ & = \frac{216(24)}{24^2 + 36} + \frac{216(6)}{6^2 + 36} + \frac{216(10)}{10^2 + 36} \\ & = \frac{5184}{612} + \frac{1296}{72} + \frac{2160}{136} \\ & = \frac{144}{17} + 18 + \frac{270}{17} \\ & = \frac{720}{17} \end{aligned}\end{equation}\tag{7}\label{eq7A}$$

Using \eqref{eq7A} above, and \eqref{eq2A}, we then have

$$\begin{equation}\begin{aligned} \frac{[XYZ]}{[PQR]} & = \frac{\left(\frac{720}{17}\right)}{240} \\ & = \frac{720}{240(17)} \\ & = \frac{3}{17} \end{aligned}\end{equation}\tag{8}\label{eq8A}$$

Finally, this gives $m=3$ and $n=17$, so $m + n = 20$.

Answer 2

Let $PZ=PX=p$, $QX=QY=q$, $RY=RZ=r$. Then $$\begin{aligned} &p+q=PX+QX=PQ=16\\ &q+r=QY+RY=QR=30\\ &r+p=RZ+PZ=RP=34\\ \end{aligned}$$

So we have $$\begin{aligned} &p=\frac{(p+q)+(r+p)-(q+r)}2=10\\ &q=\frac{(q+r)+(p+q)-(r+p)}2=6\\ &r=\frac{(r+p)+(q+r)-(p+q)}2=24\\ \end{aligned}$$

As you claimed, in order to deal with the ratio between the areas of triangles, we do not have to use "barycentric coordinates".

$$\begin{aligned} &\frac{[PXZ]}{[PQR]}=\frac{PX\cdot PZ}{PQ\cdot PR}=\frac{10^2}{16\cdot34}=\frac{25}{8\cdot17}\\ &\frac{[QYX]}{[QRP]}=\frac{QY\cdot QX}{QR\cdot QP}=\frac{6^2}{30\cdot16}=\frac{3}{5\cdot8}\\ &\frac{[RZY]}{[RPQ]}=\frac{RZ\cdot RY}{RP\cdot RQ}=\frac{24^2}{34\cdot30}=\frac{48}{5\cdot17}\\ \end{aligned}$$

Note that I write $[PQR]$ also as $[QRP]$ or $[RPQ]$ so that all three equalities are symmetric.

$$\frac{[XYZ]}{[PQR]}=1-\frac{[PZX]}{[PQR]}-\frac{[QXY]}{[PQR]}-\frac{[RYZ]}{[PQR]}=\frac{120}{5\cdot8\cdot17}=\frac3{17}$$

$m=3$, $n=17$, $m+n=20$.

---

You might wonder why $\frac{[PXZ]}{[PQR]}=\frac{PX\cdot PZ}{PQ\cdot PR}$.

Let us consider $\triangle PXZ$ and $\triangle PQZ$.
Let $ZF$ be an altitude of $\triangle PXZ$. So $[PXZ]=\frac12PX\cdot ZF$.
Since $ZF$ is also an altitude of $\triangle PQZ$, $[PQZ]=\frac12PQ\cdot ZF$.
$$\frac{[PXZ]}{[PQZ]}=\frac{\frac12PX\cdot ZF}{\frac12PQ\cdot ZF}=\frac{PX}{PQ}$$

Similarly, we have $$\frac{[PZQ]}{[PRQ]}=\frac{PZ}{PR}$$

Multiplying the two equalities above, we get $\frac{[PXZ]}{[PRQ]}=\frac{PX}{PQ}\frac{PZ}{PR}$, which is what we wanted.

Similarly, we have $\frac{[QYX]}{[QRP]}=\frac{QY\cdot QX}{QR\cdot QP}$ and $\frac{[RZY]}{[RPQ]}=\frac{RZ\cdot RY}{RP\cdot RQ}$.

Answer 3

John Omeilan has provided an excellent solution. However the answer can be simpler if we observe that $\Delta PQR$ is a right angled triangle.

Since $16^2+30^2=34^2$, $\Delta PQR$ is a right angled triangle and hence $ [PQR] = \frac{16 \times 30}{2}=240$

$s=\frac{p+q+r}{2}=\frac{30+16+34}{2}=40$

Let the radius of the inscribed circle be $x$

From $xs=[PQR] =240$, we have

$$x=\frac{240}{40}=6$$

Notice that $\sin R=\frac{16}{34}=\frac{8}{17}, \sin P=\frac{15}{17}$

and $\sin Q=\frac{34}{34}=1$

Also note that $\angle YOZ=180^o-R$.

Hence $ [YOZ]=\frac{1}{2}x^2\sin(180^o-R)=\frac{1}{2}\times 6^2 \times \sin R=18 \times \frac{8}{17}$

Similarly $[ZOX]=\frac{1}{2}x^2 \times \frac{15}{17}=18 \times \frac{15}{17}$

and $[XOY]=\frac{1}{2}x^2=\frac{1}{2}\times 6^2=18$

Hence $[XYZ]=18 \times \left(\frac{8}{17}+\frac{15}{17}+1 \right)=\frac{720}{17}$

Thus $\frac{[XYZ]}{[PQR]}=\frac{\frac{720}{17}}{240}=\frac{3}{17}$