Congruence of 3 triangles in a pyramid stereometry problem

[Ognjen]

The problem formulation:

Find the volume of three-sided pyramid SABC if SA = SB = SC = 1, <) (meaning angle) ASB = 60 deg, <)ASC = 90 deg, <)BSC = 120 deg.

By analyzing the three triangles ( ASB, ASC, BSC ) I find AB = 1, CA = 2^(1/2) and BC = 3^(1/2).

By noticing that the sum of squares of AB and CA equals the square of BC, I can apply the ''reverse Pythagorean theorem'' and deduce that <)A equals 90 deg.

However, the next part confuses me.

The official solution says the following:

Since O is the center of the circumscribed circle ( because triangle SOA is congruent with the triangle SOB and the triangle SOB is congruent with the triangle SOC, which implies that OA = OB = OC ) of the right triangle ABC ( <)A = 90 deg. ), that means that O is an element of BC and that OB = OC, so H = SO = 1/2.

H = 1/2 is clear ( it's simply deduced using trigonometry on the ( isosceles ) BSC triangle ). However, I don't quite understand how they come up with the idea that O is in the center of the line BC at the first place. Essentially, my misunderstanding comes from the fact that I don't quite understand their ''proof'' of the congruence of the three mentioned triangles. Triangles SOB ( with O being the point that divides BC in half ) and SOC have SO in common, as well as the lines SB and SC ( the equivalence of those two is given in the text of the problem ), as well as another thing which I can't determine properly: to me, it's ''obvious'' that both the two angles at O are 90 deg. and that BO = OC, but congruence of triangles only requires 3 ''components''; this means that I'm ''jumping to conclusions'' too soon and in a mathematically unjustified manner, right ? Either the right angles at O, or the equivalence of OB and OC is the RESULT of congruence of the two triangles, and not the condition under which they are congruent.

But the practically more troubling bit comes with establishment of relations between the two mentioned triangles and ASO. I can see that SO is one condition of congruence and SA the other ( its equivalence to SB and SC is, again, given in the problem ). However, I have absolutely no idea what the third could be. In the picture of the pyramid, they drew a right angle at O ( <)AOS ), which ( to my mind ) implies that this is the third condition of congruence. However, I can't at all see why this would be.

For clarity, I will post the drawing the provided in the official solution, but to me, it was of no use apparently.

The angle <) ASO was drawn by me, it isn't contained in the drawing itself.

I would immensely appreciate any kind of help with this.

 

[Dr.Peterson]

The official solution says the following:

Since O is the center of the circumscribed circle ( because triangle SOA is congruent with the triangle SOB and the triangle SOB is congruent with the triangle SOC, which implies that OA = OB = OC ) of the right triangle ABC ( <)A = 90 deg. ), that means that O is an element of BC and that OB = OC, so H = SO = 1/2.

H = 1/2 is clear ( it's simply deduced using trigonometry on the ( isosceles ) BSC triangle ). However, I don't quite understand how they come up with the idea that O is in the center of the line BC at the first place. Essentially, my misunderstanding comes from the fact that I don't quite understand their ''proof'' of the congruence of the three mentioned triangles. Triangles SOB ( with O being the point that divides BC in half ) and SOC have SO in common, as well as the lines SB and SC ( the equivalence of those two is given in the text of the problem ), as well as another thing which I can't determine properly: to me, it's ''obvious'' that both the two angles at O are 90 deg. and that BO = OC, but congruence of triangles only requires 3 ''components''; this means that I'm ''jumping to conclusions'' too soon and in a mathematically unjustified manner, right ? Either the right angles at O, or the equivalence of OB and OC is the RESULT of congruence of the two triangles, and not the condition under which they are congruent.

You haven't said how they define O, but presumably it is the orthogonal projection of S on plane ABC, and therefore it is the circumcenter of triangle ABC. This is what they are proving: the triangles SOA, SOB, SOC are congruent by the HL theorem; that is, they are right triangles with side SO shared, and hypotenuses all having the same length, 1. This implies that OA = OB = OC, which defines the circumcenter; and the circumcenter of a right triangle is the midpoint of the hypotenuse.

But the practically more troubling bit comes with establishment of relations between the two mentioned triangles and ASO. I can see that SO is one condition of congruence and SA the other ( its equivalence to SB and SC is, again, given in the problem ). However, I have absolutely no idea what the third could be. In the picture of the pyramid, they drew a right angle at O ( <)AOS ), which ( to my mind ) implies that this is the third condition of congruence. However, I can't at all see why this would be.

Have I answered your question? The HL theorem is a special case of SSA, which normally is not valid, but works when the angle is a right angle.

The picture is probably part of the trouble; by showing O on BC, rather than first prove that it is the circumcenter (OA = OB = OC), and then show that this lies on BC, they have invited your confusion. It doesn't help that they didn't state what theorem they were using to prove congruence.

 

[Ognjen]

You haven't said how they define O, but presumably it is the orthogonal projection of S on plane ABC, and therefore it is the circumcenter of triangle ABC. This is what they are proving: the triangles SOA, SOB, SOC are congruent by the HL theorem; that is, they are right triangles with side SO shared, and hypotenuses all having the same length, 1. This implies that OA = OB = OC, which defines the circumcenter; and the circumcenter of a right triangle is the midpoint of the hypotenuse.


Have I answered your question? The HL theorem is a special case of SSA, which normally is not valid, but works when the angle is a right angle.

The picture is probably part of the trouble; by showing O on BC, rather than first prove that it is the circumcenter (OA = OB = OC), and then show that this lies on BC, they have invited your confusion. It doesn't help that they didn't state what theorem they were using to prove congruence.

I didn't know that the circumcenter of a right triangle is always in the middle point of its hypothenuse. Is there any simple way to prove this, so that it's ''cleared out'' for me and not learned by heart for practical reasons ?

Thank you for clarification of what I got right and wrong. Having the HL theorem in mind will be handy in the future

 

[blamocur]

I didn't know that the circumcenter of a right triangle is always in the middle point of its hypothenuse. Is there any simple way to prove this, so that it's ''cleared out'' for me and not learned by heart for practical reasons ?

The angle between two connected segments of a circle is always one half of the opposite arc.

 

[Dr.Peterson]

I didn't know that the circumcenter of a right triangle is always in the middle point of its hypothenuse. Is there any simple way to prove this, so that it's ''cleared out'' for me and not learned by heart for practical reasons ?

Thank you for clarification of what I got right and wrong. Having the HL theorem in mind will be handy in the future

Draw a right triangle, and show that the midpoint of the hypotenuse is equidistant from the three vertices, perhaps using similar or congruent triangles. It's a good exercise, and not too hard.

Or, do what @blamocur just said, and think in terms of an angle inscribed in the circumcircle. It can be done several ways.

 

[The Highlander]

I didn't know that the circumcenter of a right triangle is always in the middle point of its hypothenuse. Is there any simple way to prove this, so that it's ''cleared out'' for me and not learned by heart for practical reasons ?

If you consider any circle with diameter AC then, by definition, the midpoint of AC (O) is the centre of the circle.

​

If you now pick any point (B) on the circumference of the circle then ∠ABC = 90°.

​

Hopefully, you may already be aware of this fact but, if not, please visit the website: Circle Theorems and scroll down to: "Angle in a Semicircle (Thales' Theorem)" for confirmation (& further explanation). It's probably worth a visit anyway to 'refresh' your "knowledge" of circles (and further explanation of blamocur's point.)

Can you now "see" that if you circumscribe any right-angled triangle then the midpoint of its hypotenuse is the centre of the circle circumscribing it?

I trust that clears up at least some of your confusion? ?

PS: If the two diagrams don't appear just come back later (when imgbb.com is working again?).

 

[Ognjen]

If you consider any circle with diameter AC then, by definition, the midpoint of AC (O) is the centre of the circle.

​

If you now pick any point (B) on the circumference of the circle then ∠ABC = 90°.

​

Hopefully, you may already be aware of this fact but, if not, please visit the website: Circle Theorems and scroll down to: "Angle in a Semicircle (Thales' Theorem)" for confirmation (& further explanation). It's probably worth a visit anyway to 'refresh' your "knowledge" of circles (and further explanation of blamocur's point.)

Can you now "see" that if you circumscribe any right-angled triangle then the midpoint of its hypotenuse is the centre of the circle circumscribing it?

I trust that clears up at least some of your confusion? ?

PS: If the two diagrams don't appear just come back later (when imgbb.com is working again?).

Thank you so much for your answer to my question, and even more so for this amazing resource ! I will make good use of it