5, looks like 5 = the radius of the semicircle

... solved with some tedious trigonometry coupled with algebraic manipulations etc.

... perhaps there is an elegant geometric solution ...

My approach was also seemingly complex evaluating the relevant triangles using Pythagoras (see attached diagram). d = diameter and r = radius. Simple inspection shows the crucial last step to be:

√[(2√5)2 - √(d2 – 62)/2)2] = (d – 6)/2

Squaring leads to 20 – [(d2 – 36)/4] = [(d – 6)/2]2 which looks to be a pain…

Or 20 – [(d2 – 36)/4] = [d2 – 12d + 36]/4; then multiplying by 4 leads to:

80 - d2 + 36 = d2 – 12d + 36 or 2d2– 12d – 80 = 0 → d2 – 6d – 40 = 0

With solution (d – 10)*(d + 4) = 0 and d either 10 (i.e. r = 5) or d = - 4 (ruled out although mathematically possible)

Care is needed in both cases of taking diameters (= 2r) or radii (= d/2).

I feel, too, there’s a simpler (more elegant?) route perhaps involving similar/congruent triangles leading to r = 5 also.

The real funny thing was that my hand-drawn diagram was initially not permitted to be attached as a .png or a .gif. I finally attached it as a .jpeg. Something about publisher's rights (yes, it's my diagram!).

Another thought involves assuming that the solution is an integral number and that ACD is the right-angled triangle with one side (not the hypotenuse) 6 units. Obviously, the other side can only be 8 units and the last side (hypotenuse) - the required diameter - 10 units. That is, it’s a simple 3, 4, 5 triangle. Such an assumption can’t be considered if there’s a more complex solution. Then, x = AP = PC = 8/2 or 4 units. And, y = 2 units from the 2root5 and 4 sided triangle. And r - y, 3 units. So, 3, 4, 5 triangles in the set up.

Peter Breuer I also thought of routing via the sine rule but was too deeply engrossed in complex math by then!

... an attempt for short solution...

In light of Alan F Rawle diagram

https://www.researchgate.net/profile/Alan-Rawle/post/Who_will_attempt_to_solve_the_new_problem_and_evaluate_the_authors_idea2/attachment/656e4e879141d41f472fe53b/AS%3A11431281209240745%401701727298231/image/Math+Diagram+JPEG.jpg?_tp=eyJjb250ZXh0Ijp7ImZpcnN0UGFnZSI6InByb2ZpbGUiLCJwYWdlIjoicXVlc3Rpb24ifX0

2 relations can be derived:

1st, PO = DC/2 = 3 because PO is a midsegment in triangle ADC.

Hence, BP = r-3. And,

2nd, triangle BPA is similar to triangle ABD because both triangles share same angles.

Hence, proportion BP-to-AB = is equal to proportion = BA-to-AD.

Thus, equality BP/AB = BA/AD yields (r-3)/(2sqrt(5)) = 2sqrt(5)/(2r),

which is simplified to r(r-3)=10.

Another thought occurred to me with respect to the 2 small triangles, ABP and BPC, that have 2√5 (or √20) as hypotenuse. Again, assuming integral answers, we have to find the other 2 sides that their squares sum to 20. This can only be 16 and 4 i.e. the other sides are 4 and 2 units in length. Thus, AP + PC = AC = 8 units. We then solve for the hypotenuse of the large triangle ACD leading to AD = 10 units (diameter) and thus r = 5 units.

Steftcho P. Dokov That's the elegant solution I think! Well done!

Liudmyla Hetmanenko Alan F Rawle Steftcho P. Dokov

step by step solution.

1. Draw the semicircle with diameter AD.

A______D

| |

| |

| |

|_______|

B C

2. Since AB = BC = 2√5, we can conclude that triangle ABC is an isosceles triangle.

3. Draw the altitude from point B to side AC, and let's call the point of intersection E.

A______D

| |

| |

| E |

|_______|

B C

4. Triangle ABC is isosceles, so the altitude from point B to side AC bisects AC into two equal segments.

5. Therefore, AE = EC = √5.

6. Now, we have a right-angled triangle ABE, where AB = 2√5 and AE = √5.

7. Using the Pythagorean theorem, we can find the length of BE.

BE^2 = AB^2 - AE^2

BE^2 = (2√5)^2 - (√5)^2

BE^2 = 20 - 5

BE^2 = 15

Taking the square root of both sides:

BE = √15

8. Since BE is the radius of the semicircle, the radius of the semicircle is √15.

Therefore, the radius of the semicircle is √15.

Hope it helps.

Mohammad Imam

It's always useful to have a diagram - I simply made mine by adapting that of Liudmyla Hetmanenko

I think your Point E is 'my' point P in my diagrams above.

Consider your point/step 5. Yes, ABC is an isosceles triangle, but it's not an equilateral triangle. In your notation, AE = EC yes, but they are 4 units each (the triangles ABE and EBC have hypotenuse 2√5 and the other sides are BE = 2 and EC = 4 units in length).

Mohammad Imam, briefly, everything is okay only in the first 4 steps in your writings. However, step 5 makes an unjustifiable assertion, and so there is a wrong flow in the derivations until the last step 8. Hopefully, you may notice that your answer differs from the other approaches answer. Probably, you have a chance 1-to-4 for being right = 1 = your answer = vs. = 4 other solutions with another answer. And, 1/4 = 25% is a small chance after all.

if phi is half of angles AOB or BOC , then half of COD angles are 90-2phi degrees. Half of Triangle BOC cut thru O gives R=sqrt(5)/sin(phi) and half of COD cut thru O gives R= 3/cos (2phi). Now, cos 2(phi)=1-sin^2 (phi), so sin phi =sqrt(5)/R, cos 2(phi)= 1-(10/R^2); R=3R^2/(R^2-10); R^2-3R-10=0, (R-5)(R+2)=0; since R cannot be negative, R=5

Alan F Rawle Steftcho P. Dokov Sumit Bhowmick Peter Breuer Mohammad Imam

Dear colleagues, thank you for your expressed interest and active discussions.

I am delighted to see that these problems are of interest not only to me.

I would like to share my ideas on solving this particular problem.

The deltoid ABCD, formed by two isosceles triangles ABC and AOC, suggests the application of the Lagrange formula for the bisectors OP and BP (AC∩BO=P) or the Pythagorean theorem for triangles ABP and AOP.

I am grateful to everyone for their contributions to this work.

Liudmyla Hetmanenko

Hello Liudmyla! All problems you posted are very nice! If you have another solution for above problem, not using trigonometry, please post it! At the moment the only pure geometrical solution is the very nice solution of S.O.Dokov.

Another solution( not pure geometrical) can be:

Let t be the measure af angle ADC. ACD is a right angle, then

tan t =(sin t)/(cos t)=AC/6 => 36sin2t =AC2 cos2t

AC2=(2√5)2+(2√5)2-40 cos(1800-t) = 40(1-cos(1800-t))=40(1+cos t)

Therefore 36 sin2t =40(1+cos t) cos2t => 36-36 cos2t = 40cos2t +40cos3t

So we obtain 40cos3t+76cos2t -36=0, equivalently to

10cos3t+19cos2t -9=0

The equation 10x3+19x2-9=0 has -1 as solution

10x3+19x2-9= 10x3+10+19x2-19 = (X+1)(10X2-10X+10 +19X-19).

The equation 10x2+9x-9=0 has solutions x=-3/2 and x=3/5

The valid solution is x=cos t=3/5

But cos t= CD/AD=6/AD => 6/AD=3/5, then AD = 10

I will call x half of angle ABC.

We can compute AC^2 from right triangle ACD and from isosceles triangle ABC to get the equation and we equal the two forms to get

4 r^2 - 36 = 80 sin^2 x = 80 sin^2 BAO because triangle BAO is isosceles.

From the right triangle BAD, sin^2 BAO = (4 r^2 - 20)/4r^2

If we plug in above we get

r^2 - 29 r^2 + 100 = 0, with solutions r = 5 and r = 2. The second is obviously not good.

Liudmyla Hetmanenko Especially for you, another problem about inscribed quadrilaterals:

Let ABCD be a quadrilateral inscribed in a circle, having orthogonal diagonals. If P is the intersection point of diagonals AC and BD, then

PA*PC

Dinu Teodorescu, thanks a lot for pointing out my non-trigonometry solution.

Another such not-using-trigonometry solution is the next one.

Again in light of Alan F Rawle diagram:

we have that PO = DC/2 = 3 because PO is a midsegment in triangle ADC.

Hence, BP = r-3.

Now, denote the length of AP by x, and write Pitagorian theorem two times.

1st for triangle APB: x^2 + (r-3)^2 = 20

and

2nd, for triangle AOP: x^2 + 9 = r^2

Thus, solve the system with two equations for x-and-r, and that's it.

Good luck

Sorry for mistake! In my above solution AB must be replaced by AC( I do it)! Now all is ok!

Another mistake relatively to the problem posted by me!

The conclusion must be PA*PC

Dinu Teodorescu, about your question

********

Let ABCD be a quadrilateral inscribed in a circle, having orthogonal diagonals. If P is the intersection point of diagonals AC and BD, then PA*PC

Steftcho P. Dokov Congrats Steftcho, very nice solution!

Another solution:

Clearly, triangles PAB and PDC being similar, we have PA*PC=PB*PD

Now we use the simplest version of Cauchy - Schwarz inequality, i.e. the version in R2:

(ax+by)2