# Points on the hypotenuse of a right-angled triangle

Points $$K$$ and $$L$$ are chosen on the hypotenuse $$AB$$ of triangle $$ABC$$ $$(\measuredangle ACB=90^\circ)$$ such that $$AK=KL=LB$$. Find the angles of $$\triangle ABC$$ if $$CK=\sqrt2CL$$.

As you can see on the drawing, $$CL=x$$ and $$CK=\sqrt2x$$.

I don't know how to approach the problem at all. Since $$\measuredangle ACB=90^\circ$$, it will be enough to find the measure of only one of the acute angles. If $$\measuredangle ACK=\varphi_1$$ and $$\measuredangle BCL=\varphi_2$$, I have tried to apply the law of sines in triangle $$KCL$$, but it seemed useless at the end. Thank you! I would be grateful if I could see a solution without using coordinate geometry.

• Law of cosines may work. Commented Aug 11, 2021 at 16:43
• @herbsteinberg, may I ask you to be a little more specific? Commented Aug 11, 2021 at 16:46
• For each of the triangles you can write an equation for three sides and an angle. For example - middle triangle: $u^2=2x^2+x^2-2\sqrt{2}x^2cos$(b) where $u$ is $\frac{1}{3}$ hypotenuse and b is the middle piece of the right angle. You can get a total of four equations this way with four lengths $(u,x,v,w)$ (v and w sides) and three angles (a,b,c) (sum to right angle) as unknowns. Two more equations needed. Commented Aug 11, 2021 at 17:02
• I've added a comment. You can see if it actually helps. :) Commented Aug 11, 2021 at 17:58

WARNING: This solution uses co-ordinate geometry, which had not been explicitly excluded at the time of answering.

Choose $$C$$ as origin, and $$CB$$ as $$x$$-axis. Then $$CA$$ will be the $$y$$-axis.

Let $$A(0,a)$$ and $$B(b,0)$$. Then, using the well-known section formula, $$L \equiv \left(\frac {2b}{3}, \frac a3\right)$$ and $$K\equiv \left(\frac b3, \frac {2a}{3} \right)$$ Now we have: $$CK^2=2\cdot CL^2$$ Using the distance formula, this means that: $$\left(\frac b3\right)^2+\left( \frac {2a}{3} \right)^2=2\left( \frac {2b}{3} \right)^2+2\left( \frac a3 \right)^2$$ This simplifies to: $$\frac ab=\sqrt {\frac 72}$$ Thus, $$\frac {AC}{CB}=\tan \beta=\sqrt {\frac 72}$$. Hence, $$\beta=\tan^{-1} \sqrt{\frac 72}$$ and $$\alpha=\cot^{-1} \sqrt {\frac 72}$$.

• Anyways I'm not participating in this discussion anymore. OP edited the question later to exclude coordinate geometry explicitly. I'll add a warning in my answer saying it uses the same. Commented Aug 11, 2021 at 17:34

Let $$AK=KL=LB=x$$. By the definition of cosine of an acute angle $$\cos\beta=\dfrac{a}{3x}$$ By the law of cosines in triangle $$KBC$$ $$\cos\beta=\dfrac{a^2+4x^2-CK^2}{4ax}$$ By the law of cosines in triangle $$LBC$$ $$\cos\beta=\dfrac{a^2+x^2-CL^2}{2ax}$$ So $$\dfrac{a^2+4x^2-CK^2}{4ax}=\dfrac{a^2+x^2-CL^2}{2ax}$$ Using $$CK^2=2CL^2$$, this can lead you to $$\dfrac{a}{x}=\sqrt{2}$$. This means $$\cos\beta=\dfrac{\sqrt2}{3}$$.

• Your solution is the best one. I like it and have already upvoted you. Commented Aug 11, 2021 at 17:57
• Thank you!................. Commented Aug 11, 2021 at 17:59
• The original poster chose your solution and it means all you said was right. Commented Aug 11, 2021 at 18:01
• How can i find alpha, though? Commented Aug 11, 2021 at 18:02
• I'll leave that for you to answer. :) Commented Aug 11, 2021 at 18:02

Let's construct median of right triangle say, $$6y=CD$$

$$AD=BD=CD=3y$$

$$AK=KL=LB=2y$$

$$KD=DL=y$$

Applying median theorem in $$\triangle CKL$$

$$2\times (3y)^2=x^2+2x^2-\frac{(2y)^2}{2}$$

$$x=\frac{2\sqrt{15}y}{3}$$

Applying cosine theorem in $$\triangle CDL$$

$$\angle CDL=2\alpha$$

$$\cos2\alpha=\frac{(3y)^2+y^2- (\frac {2\sqrt{15}y}{3})^2}{2 \times 3y\times y} = \frac{5}{9}$$

$$2\alpha= cos^{-1}(\frac{5}{9})=56.25$$ $$\alpha=28.13$$

$$\beta=61.87$$

Or

$$\cos2\alpha = 2cos^2\alpha-1=\frac{5}{9}$$

$$\cos\alpha= \frac{\sqrt7}{3}$$

Since $$CL$$ is a median of $$\Delta KCB$$, we obtain: $$CL=\frac{1}{2}\sqrt{2CK^2+2CB^2-KB^2}$$ or $$x=\frac{1}{2}\sqrt{4x^2+2a^2-\left(\frac{2}{3}\sqrt{a^2+b^2}\right)^2}$$ or $$\frac{b^2}{a^2}=3.5$$ and $$\beta=\arctan\sqrt{3.5}.$$

Draw horizontal lines from points K and L of side $$\overline{AB}$$ to points $$K_A$$ and $$L_A$$ of side $$\overline{BC}$$. Because points $$K$$ and $$L$$ trisect side $$\overline{AB}$$, then points $$K_A$$ and $$L_A$$ trisect side $$\overline{BC}$$. So $$\triangle{CK_AK}$$ and $$\triangle{CL_AL}$$ are right triangles.

Draw vertical lines from points K and L of side $$\overline{AB}$$ to points $$K_B$$ and $$L_B$$ of side $$\overline{AC}$$. Because points $$K$$ and $$L$$ trisect side $$\overline{AB}$$, then points $$K_B$$ and $$L_B$$ trisect side $$\overline{AC}$$ So $$\triangle{CK_BK}$$ and $$\triangle{CL_BL}$$ are right triangles.

We can now argue that

\begin{align} CL &= \sqrt 2 \; CK \\ \sqrt{\left(\dfrac 23b \right)^2 + \left(\dfrac 13a \right)^2} &= \sqrt 2 \cdot \sqrt{\left(\dfrac 13b \right)^2 + \left(\dfrac 23a \right)^2} \\ a^2 + 4b^2 &= 2(4a^2+b^2) \\ a^2 + 4b^2 &= 8a^2+2b^2 \\ 2b^2 &= 7a^2 \\ b &= \sqrt{\dfrac{7}{2}} a \end{align}

Since $$\triangle{ACB}$$ is a right triangle,

\begin{align} c^2 &= a^2 + b^2 \\ c^2 &= a^2 + \dfrac 72 a^2 \\ c^2 &= \dfrac 92 a^2 \\ c &= \dfrac{3}{\sqrt 2} a \end{align}

It follows that $$a:b:c = \sqrt 2 : \sqrt 7 : 3$$.

$$m\angle A = \arcsin \dfrac{\sqrt 2}{3} \approx 28.13^\circ$$.

• Thanks! I am a little confused. You and Medi got different answers. Which are correct? Commented Aug 14, 2021 at 13:27
• @MathGuy: I just noticed that I got points K and L backwards. But the results should still be correct. I also noticed that Medi uses x for 1/3 of the hypotenuse as well as CL. I didn't check to see if that affected his answer. Commented Aug 14, 2021 at 22:09
• @MathGuy : I just went over everything again and corrected my errors. I still swapped points K and L but the numbers I used are correct for the picture you drew. It seemed to me that this problem was a good example of using "parallel lines divide transversals proportionally" so I tried to present that. It's just that my dyslexia is getting worse as I get older and I make more mistakes now. Commented Aug 19, 2021 at 20:58

Perform point-symmetry with respect to the point $$L$$. Then the image of $$C$$ is the point $$D$$ such that $$D$$ lies on the line $$CL$$ and $$CL = DL$$. By assumption, $$BL = KL$$ which means that $$K$$ is the symmetric image of $$B$$. Therefore, the quad $$BCKD$$ is by construction a parallelogram. However, the problem states the $$CD \cdot CL = 2\,CD \cdot CD = 2\,CD^2 = CK^2$$ which translates to $$\frac{CL}{CK} = \frac{CK}{CD}$$ which combined with the fact that $$\angle \, LCK = \angle \, KCD$$ implies that the triangles $$\Delta \, CKL$$ and $$\Delta \, CDK$$ are similar. Consequently, $$\angle \, CKL = \angle \, CDK = \theta$$ But since $$BCKD$$ is a parallelogram, $$\angle \, LCB = \angle \, DCB = \angle \, CDK = \theta$$ Putting these angle equalities together, we get $$\angle \, CKB = \angle\, CKL = \theta = \angle \, LCB$$ Because of the latter equality and the fact that the two triangles $$\Delta \, BCK$$ and $$\Delta \, BLC$$ share the common angle at the vertex $$B$$, they are similar triangles. Therefore $$\frac{BL}{BC} = \frac{BC}{BK}$$ or alternatively $$BK \cdot BL = BC^2$$ $$2 \, BL \cdot BL = BC^2$$ $$2 \, BL^2 = BC^2$$ Replace by $$BL = \frac{AB}{3}\,$$ which yields $$2 \, \left(\frac{AB}{3}\right)^2 = BC^2$$ So from there, we get $$\sqrt{2} \frac{AB}{3} = BC$$ which becomes $$\sin(\angle \, BAC) = \cos(\angle \, ABC) = \frac{BC}{AB} = \frac{\sqrt{2}}{3}$$ Hence

$$\angle \, BAC = \arcsin\left( \frac{\sqrt{2}}{3} \right)$$ $$\angle \, ABC = \arccos\left( \frac{\sqrt{2}}{3} \right)$$