The only isosceles triangle with a special construction

Question

An isosceles triangle $ABC$ has a base $AB$. A segment of length equal to the base $AB$ connects two points on the legs, point $P$ on $AC$ and point $Q$ on $BC$, such that the length of both segments $AP$ and $CQ$ is also equal to the base $AB$. How to calculate the angle at the vertex $C$ of this triangle?

---

Illustration mine, not super precise. Spent a couple hours trying to solve this problem from a lower high school Olympiad problem set book for my son, and despite having a Comp Sci degree and passing many math exams in my time... So far I'm only able to prove that there must be one and only one such a triangle, but can't seem to find enough independent equations for the angles. Any hints appreciated.

Answer 1

Draw a circle with center $P$ and radius $PQ$ and let it intersect $BC$ at $R$. Note the isosceles triangles and their external angles. It is seen that $$\widehat{PQR} = 2\widehat{C} $$ $$\widehat{APR} = 3\widehat{C} \qquad (1)$$ $$\widehat{PAR} = \frac12 (180^o - 3\widehat{C})$$ $$\widehat{RAB} = \widehat{C} \qquad (2)$$ From (2) we conclude that $\triangle {RBA}$ is similar to $\triangle {ABC}$ and is therefore isosceles, which means $AB = AR$. But then that means $\triangle {APR} $ is equilateral and its angles are $60^o$. Finally, from (1) we find that $\widehat{C} = 20^o$.

Answer 2

$\text{\{length of green line segment\}}=\dfrac12\text{\{length of red line segment}\}$

(This is obvious if you slide the green line segment left so that it is half the triangle's base.)

$\therefore 3\theta-180^\circ=60^\circ$

$\therefore \theta=80^\circ$

So the angle at the top vertex is $20^\circ$.

Answer 3

Draw $\overline{AQ}$. Use the equality of base angles in an isosceles triangle to show that $\angle BQA$ measures $3/2$ times the apex angle $\theta$ you are trying to find, and the base angles of isosceles $\triangle AQP$ measure $\theta/2$. With this information you should be able to find the ratio of $|\overline{AQ}|$ to $|\overline{AB}|$ and then set up the Law of Sines in $\triangle ABQ$ to get an equation for $\theta$.

The resulting equation looks complicated at first but is in fact simple to solve. While there are indeed two solutions for $\theta$ between $0°$ and $180°$, only one is small enough to match the geometry you draw. The larger root won't have $P$ between $A$ and $C$, among other things.

Answer 4

We start with the given $\Delta ABC$, which is isosceles, $CB=CA$, and two points on the sides, $P\in CB$, $Q\in CA$ as given, $CQ=QP=PB=BA$ and we may and do assume that the common length is the unit $1$. Denote such a constellation of points by $\maltese(CQPBA)$, let us call it an $L$-configuration (where $L$ is a mnemonic for Langley).

---

We copy paste the same triangle to obtain an isosceles triangle $\Delta BCD$ as in the figure, so that denoting by a prime the reflection w.r.t. the perpendicular bisector of $BC$ we have $A'=D$ and $B'=C$. Let $R=P'\in BC$, $S=Q'\in BD$. Consider also $T\in CA$, $U\in BD$, (intermediate points on segments,) so that $AT=DU$ are also the unit. Then there are the following $L$-configurations in the picture:

• $(i)$ $\maltese(CQPBA)$, the given $L$-configuration,
• $(ii)$ $\maltese(CRTAB)$, which is $(i)$ reflected w.r.t. symmetry axis of $\Delta CBA$,
• $(iii)$ $\maltese(BSRCD)$, which is $(i)$ reflected w.r.t. symmetry axis of $BC$,
• $(iv)$ $\maltese(BPUDC)$, which is $(iii)$ reflected w.r.t. symmetry axis of $\Delta BCD$.

---

Here are some simple observations. Let $x$ denote the wanted $\hat C$-angle in $\Delta ABC$. $ABCD$ is an isosceles trapezium, $AD\|BC$, and there are $x$-angles in each vertex $A,B,C,D$ built by one of the parallel sides and a diagonal. $\Delta BSR$ is isosceles, so it has two $x$-angles in $B,R$. In particular the segments $CQ,RS$ are parallel and of length one. Then the parallelogram $CRSQ$ is a rhombus, because the side $CR$ is also of length one. The point $S$ is thus on the bisector of its angle $\hat C$, same angle as in $\Delta ABC$, so this bisector is the symmetry axis of the triangle. In particular $SA=SB$ is also the unit length.

---

So $\Delta SAB$ is equilateral.

---

Now it is easy to conclude. For instance observe that $\Delta QSA$ is a translated version of $\Delta CRT$, so $x=\widehat {QAS}=\widehat {AQS}$, making $\widehat {QSA}=180^\circ-2x$. By symmetry, $\widehat {RSB}=180^\circ-2x$.

We are counting the angles around $S$ and obtain: $$ 360^\circ = (180^\circ-2x)+x+(180^\circ-2x)+60^\circ\ .$$ So $3x=60^\circ$, and finally $\bbox[lightgreen]{\ x=20^\circ\ }$.

$\square$

---

In the final picture there is a bonus property, the circle $SPRQUT$. To see that the six points are on a circle, hold $Q,R$ fixed and observe that there are three $x$-angles: $x=\widehat{QTR}=\widehat{QSR}=\widehat{QPR}$. So $QTSPR$ cyclic. By symmetry, $U$ is also on this circle.

Answer 5

For any quadrilateral $XYZW$ whose diagonals $XZ,YW$ intersect at a point $D$, the sine law tells us that: $$\frac{\sin\angle DXW}{|DW|}\,\,\frac{\sin\angle DWZ}{|DZ|}\,\,\frac{\sin\angle DZY}{|DY|}\,\,\frac{\sin\angle DYX}{|DX|}\\=\frac{\sin\angle DWX}{|DX|}\,\,\frac{\sin\angle DZW}{|DW|}\,\,\frac{\sin\angle DYZ}{|DZ|}\,\,\frac{\sin\angle DXY}{|DY|} $$

$${}$$ $\qquad\qquad\qquad\qquad$

Clearing denominators we get: $$\sin\angle DXW\,\,\sin\angle DWZ\,\,\sin\angle DZY\,\,\sin\angle DYX\\=\sin\angle DWX\,\,\sin\angle DZW\,\,\sin\angle DYZ\,\,\sin\angle DXY \qquad (1)$$

In particular, for the quadrilateral $ABQP$, we have both $\bigtriangleup APB$ and $\bigtriangleup APQ$ isosceles, so $\angle ABP=\angle APB$ and $\angle PAQ =\angle PQA$.

$${}$$ $\qquad\qquad\qquad\qquad$

Thus $(1)$ simplifies to: $$\sin \angle BAQ \,\,\sin \angle QBP \,\,=\,\,\sin \angle BPQ \,\,\sin \angle AQB\qquad (2)$$

Denote our desired angle $\angle ACB=4\theta$. Then standard properties of the angles of isosceles triangles tell us that:

$$ \angle BAQ= 90^\circ -4\theta,\,\, \angle QBP=45^\circ-3\theta ,\,\, \angle BPQ=135^\circ-5\theta ,\,\,\angle AQB=6\theta.$$

Thus we may rewrite $(2)$: $$\sin (90^\circ -4\theta) \,\,\sin (45^\circ-3\theta) \,\,=\,\,\sin (135^\circ-5\theta) \,\,\sin (6\theta)\qquad (3)$$ which simplifies to:$$\cos (4\theta) \,\,\sin (45^\circ-3\theta) \,\,=\,\,\cos (5\theta-45^\circ) \,\,\sin (6\theta)\qquad (4)$$

Using the trigonometric identity: $$2\sin A\cos B =\sin(A+B)+\sin(A-B)$$ we get from $(4)$: $$\sin(45^\circ+\theta)+\sin(45^\circ-7\theta)=\sin(11\theta-45^\circ)+\sin(\theta+45^\circ)$$

Finally, cancelling the common term we have:$$\sin(45^\circ-7\theta)=\sin(11\theta-45^\circ).$$

There are two possibilities:

1. $45^\circ-7\theta=180^\circ-(11\theta-45^\circ)+360^\circ k$, for $k$ an integer.

In this case $\angle ACB=4\theta=180^\circ (2k+1)$, which cannot be the angle of a triangle. This just leaves:

2. $45^\circ-7\theta=11\theta-45^\circ+360^\circ k$, for $k$ an integer.

In this case $18\theta=90^\circ-360^\circ k$, so $\angle ACB=4\theta=20^\circ-80^\circ k$.

The only possible values for $\angle ACB$ that can occur in a triangle are therefore $20^\circ$ or $100^\circ$. However the latter is obtuse, meaning $AB$ is longer than $AC$, so the desired construction is not possible.

We conclude $\angle ACB=20^\circ$.