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Background: this problem was from social app, but I cannot find original source. This is an interesting problem that I have not seen similar ones. It is a hard problem, cannot be solved with angle chasing. I worked out via analytic geometry programming, but cannot figure out a logical approach. Is there a logical approach to solve it? May need help from some theorems that I am not familiar with.

Problem: There are 5 known angles in the figure. Find the angle x. Is there an approach to solve it through deduction?

enter image description here

r ne
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  • Yes. Apply sine/cosine/tangent rule to every triangle in sight. – user10354138 Feb 28 '21 at 00:34
  • What's wrong with plopping everything into the coordinate plane and solving for the last point? – Joshua Wang Feb 28 '21 at 00:49
  • @JoshuaWang Yeah can is solvable but not practically by hand. The answer is an integer in degrees, so there should be a logical approach. – r ne Feb 28 '21 at 01:49
  • Solving analytically, I get $x\approx 41^\circ$ and I'm willing to believe it's exactly $41^\circ$ for the moment. The only thing I see is that if we extend the bottom edge in the figure, we get an angle of $57^\circ$ to the right of $x$. This is the same as the third angle of the triangle with angles $49^\circ$ and $74^\circ=44^\circ+30^\circ$ at the bottom of the figure. So, if there's some way to deduce that these angles must be equal, we're done. I don't see any reason at all, though. Looks like a coincidence to me, but what do I know? – saulspatz Feb 28 '21 at 03:15
  • @saulspatz Yeah, analytically this answer is right. I just cannot imagine how an integer could be got without a logical reason. This should not be a calculation. – r ne Feb 28 '21 at 04:26
  • Can you prove it's exactly an integer? I took a stab at it, but gave up as the trig formulas got too complicated. BTW, it seems that not all the angles in the diagram are integers, unfortunately. In the upper-left hand corner, there's an angles or $19^\circ$, but the angle next to it is $\approx56.073^\circ$ according to my calculations. – saulspatz Feb 28 '21 at 04:29
  • @saulspatz Have you drawn this figure in GeoGebra? If so, can you share the link so that we can play on the diagram. It will take some time to reproduce the figure from scratch. – krazy-8 Feb 28 '21 at 04:38
  • @dodoturkoz No, I haven't, sorry. I printed the diagram in the question. – saulspatz Feb 28 '21 at 04:43
  • @rne Have you drawn this figure in GeoGebra? Sorry saulspatz, I didn't mean to tag you. – krazy-8 Feb 28 '21 at 05:02

1 Answers1

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Label the bottom two points (starting from left) with $A$, $B$, and the top points (starting from left) with $C$, $D$, $E$. Let $F$ be the intersection of $AD$ and $BC$. Let $\omega$ be the circumcircle of $ABF$. Let $G$ be the second common point of $\omega$ and $AE$. Let $H$ be the second common point of $BD$ and $\omega$. Finally, let $I$ be the point on the ray $BA^\to$ such that $AG=AI$.

Since $AI=AG$, we have $\angle AIG = \frac 12 \angle BAG = \frac 12 \cdot 44^\circ = 22^\circ$. Then we find that $\angle GBI = \angle FBI + \angle GBF = 49^\circ + \angle GAF = 49^\circ + 30^\circ = 79^\circ$, and $\angle IGB = 180^\circ - \angle BIG - \angle GBI = 180^\circ - 22^\circ - 79^\circ = 79^\circ$. Hence $IBG$ is isosceles and $IB=IG$.

Let $\Omega$ be the circle with center $I$ and radius $IB=IG$. Let $C'\in\Omega$ be such that the chord $BC'$ passes through $F$. Note that $\angle GIC' = 2 \angle GBC' = 2\angle GBF = 2\cdot 30^\circ = 60^\circ$, hence triangle $GIC'$ is equilateral. In particular, $C'I=C'G$. Since $AI=AG$, it follows that $AC'$ bisects the angle $GAI$. Hence $\angle GAC' = \frac 12 \angle GAI = \frac 12 \cdot (180^\circ - \angle BAG) = 90^\circ - \frac 12 \cdot 44^\circ = 90^\circ - 22^\circ = 68^\circ = \angle GAC$, and therefore $C'$ lies on the line $AC$.

Moreover, $\angle C'BI = 90^\circ - \frac 12 \angle BIC' = 90^\circ - \frac 12(\angle BIG + \angle GIC') = 90^\circ - \frac 12 \cdot (22^\circ + 60^\circ) = 49^\circ = \angle CBI$, and therefore $C'$ lies on the line $BC$.

Hence $C'$ is the intersection of $AC$ and $BC$, hence $C'$ coincides with $C$.

Next, we shall prove that $C, G, H$ are collinear. Indeed, $\angle CGH = \angle CGA + \angle AGH = \angle AIC + (180^\circ - \angle HBA) = \angle GIC + \angle BIG + 180^\circ - 82^\circ = 60^\circ + 22^\circ + 180^\circ - 82^\circ = 180^\circ$.

Applying Pascal theorem for (degenerated) hexagon $FBBHGA$ we see that $BE$ is tangent to $\omega$. Therefore $\angle EBD = \angle BAH = \angle BAF - \angle HAF = 74^\circ - \angle HBF = 74^\circ - 33^\circ = 41^\circ$.

Remark: This is a special case of the following phenomenon:

The proof presented above can be adapted to the general case.

timon92
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  • Thanks! I think this is correct. When you said hexagon FBBHGA you meant ABBHGF – r ne Feb 28 '21 at 11:52
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    @rne No, I meant $FBBHGA$. Opposite sides of this hexagon intersect at collinear points $C=FB\cap HG$, $E':=BB \cap GA$, and $D=BH \cap AF$. We see that $E'$ coincides with $E$ as both points lie on the lines $CD$ and $AG$. In other words, $BE$ is tangent to $\omega$. – timon92 Feb 28 '21 at 12:16
  • Impressive solution – saulspatz Feb 28 '21 at 12:54