Korean middle school math Question #29

Korean middle school math Question.

Hello. I am Teacher JoJo, teaching mathematics in South Korea.

Today, I have brought a math problem related to the incenter of a triangle.

This problem can be solved by 2nd-year middle school students in Korea.

Why don’t you give it a try as well?

[Korean middle school math question]

Question Description)

In triangle ABC, point I is the incenter.

\angle AEI = x, \; \; \angle BDI = y, \; \; \angle ACB = 40^\circ

\angle x + \angle y = ?

If you are familiar with the properties of the incenter of a circle, it’s definitely worth a challenge.

Try to solve it on your own first, and then compare your solution with mine below.


Have you found the answer yet?

If you’re a diligent Korean student, you’ll probably solve it in under a minute.

Let’s quickly solve it using the approach of a Korean student.

\angle ACD = 40^\circ \longrightarrow \angle AIB = 110^\circ

\angle AIB = \angle DIE = 110^\circ

\angle BEC = 180 -x, \; \; \angle ADC = 180 -y

The sum of the angles in \Box IDCE = 360^\circ

so, 40 + 180 -x + 180 -y + 110 = 360

150 = x + y

The answer is {\color{red} 150^\circ}

Did you understand my solution process?

If you are familiar with the properties of the incenter, you could have solved it quickly using this method.


if you are not aware of the properties of the incenter, this part might have been confusing for you.

{\color{blue} \angle ACD = 40^\circ \longrightarrow \angle AIB = 110^\circ}

Why does it suddenly become like this? Let me explain.

According to the properties of the incenter,

If \angle A = x^\circ, \angle BIC = \cfrac{x}{2} + 90^\circ


The incenter is the point of intersection of the angle bisectors.

Extend segment AI, and consider triangles ABI and ACI.

Let’s denote the exterior angles of each triangle.

Please refer to the diagram below.

\angle IAB + \angle IBA = \angle BID

\angle IAC + \angle ICA = \angle CID

\angle a + \angle b + \angle c = 90^\circ

\angle BIC = {\color{blue} a + b + c} + {\color{green}a} = {\color{blue} 90^\circ} + {\color{green} \cfrac {\angle A}{2}}

\therefore \angle BIC = \cfrac{\angle A}{2} + 90^\circ

Did my explanation make sense to you?

Even middle school students in Korea find this concept challenging.

If it’s difficult, please read it again.

If you take your time to examine it, anyone can understand it.

I have other Korean math problems on my website, and I also have a YouTube channel.

A Korean solving American high school math questions. #28


[Korean ver]

안녕하세요. 저는 한국에서 수학을 가르치고있는 조조쌤입니다.

오늘은 중학교 2학년 수준의 문제를 가지고왔습니다.

내심을 잘 이해하고있다면 쉽게 풀수있으리라 생각됩니다.

함께 문제 보실까요?


천천히 풀어보시고 아래의 제 풀이와 비교해 보세요.


아마 공부를 좀 하는 한국 학생이라면, 1분안에 풀었을 것같네요.

그럼 한국 학생처럼 빠르게 풀어볼까요?

\angle ACD = 40^\circ \longrightarrow \angle AIB = 110^\circ

\angle AIB = \angle DIE = 110^\circ

\angle BEC = 180 -x, \; \; \angle ADC = 180 -y

사각형IDCE의 내각의 합은 360입니다.


40 + 180 -x + 180 -y + 110 = 360

150 = x + y


{\color{red} 150^\circ}입니다.

제 풀이 이해가 되셨나요?

이해가 되었다면 내심의 성질에 대해서 잘 알고계신것입니다.

만일 그게 아니라면, 좀 더 자세한 풀이를 보여드리겠습니다.

아마 이부분이 이상했을 것입니다.

{\color{blue} \angle ACD = 40^\circ \longrightarrow \angle AIB = 110^\circ}

이것에 대해서 자세히 설명해드리겠습니다.

내심의 성질에 의하여,

만일 \angle A = x^\circ라면, \angle BIC = \cfrac{x}{2} + 90^\circ입니다.

내심의 성질 기억나나요? 증명까지 해보죠.


내심은 각의 이등분선의 교점입니다.

선분AI에 연장선을 긋습니다.

그리고 삼각형ABI, ACI에서 각각의 외심을 봅니다.

아래 그림을 보세요.

\angle IAB + \angle IBA = \angle BID

\angle IAC + \angle ICA = \angle CID

\angle a + \angle b + \angle c = 90^\circ

\angle BIC = {\color{blue} a + b + c} + {\color{green}a} = {\color{blue} 90^\circ} + {\color{green} \cfrac {\angle A}{2}}

\therefore \angle BIC = \cfrac{\angle A}{2} + 90^\circ

증명이 이해가되나요? 아마 학교에서도 배웠을 것입니다.

보통 배워도 까먹죠.

한번보고 이해가 안될수도있습니다.

천천히 다시한번 읽어본다면 더 잘 이해가 될것입니다.

그럼 오늘은 여기까지 하도록 하겠습니다.

다음에 또 재미있는 문제로 찾아뵙겠습니다.






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