Maths Olympiad Prep

Track / Stage 7 / 266 of 300 #1666 of 1964

Problem 1666

National Olympiad second round; IMO P1/P4
Geometry Difficulty 7.5 Find the answer

Giiven ΔABC\Delta ABC, ∠CAB=75∘\angle CAB=75^{\circ} and ∠ACB=45∘\angle ACB=45^{\circ}. BCBC is extended to TT so that BC=CTBC=CT. Let MM be the midpoint of the segment ATAT. Find ∠BMC\angle BMC.

The source for this one didn't record the answer, so there is nothing to check what you type against. Work it on paper and mark yourself against the solution below.

Next problem →

Official solution

1. Given ΔABC\Delta ABC with ∠CAB=75∘\angle CAB = 75^\circ and ∠ACB=45∘\angle ACB = 45^\circ. We need to find ∠BMC\angle BMC where BCBC is extended to TT such that BC=CTBC = CT and MM is the midpoint of ATAT.
2. Draw the altitude from vertex AA to BCBC and call the intersection point HH. This gives us two right triangles ΔABH\Delta ABH and ΔAHC\Delta AHC.
3. Since ∠CAB=75∘\angle CAB = 75^\circ and ∠ACB=45∘\angle ACB = 45^\circ, we can find ∠BAH\angle BAH and ∠HAC\angle HAC:
∠BAH=75∘−45∘=30∘ \angle BAH = 75^\circ - 45^\circ = 30^\circ
∠HAC=45∘ \angle HAC = 45^\circ
4. Let BH=xBH = x. Then, using trigonometric ratios in ΔABH\Delta ABH and ΔAHC\Delta AHC:
AB=2x(since sin⁡30∘=12) AB = 2x \quad (\text{since } \sin 30^\circ = \frac{1}{2})
AH=x3(since tan⁡30∘=13) AH = x\sqrt{3} \quad (\text{since } \tan 30^\circ = \frac{1}{\sqrt{3}})
HC=x3(since tan⁡45∘=1) HC = x\sqrt{3} \quad (\text{since } \tan 45^\circ = 1)
AC=x6(using Pythagoras’ theorem in ΔAHC) AC = x\sqrt{6} \quad (\text{using Pythagoras' theorem in } \Delta AHC)
5. Since BC=CTBC = CT, AM=MTAM = MT implies MM is the midpoint of ATAT. Therefore, MCMC is the midsegment of △ABT\triangle ABT and MC=xMC = x.
6. BMBM and ACAC are medians. Let KK be the intersection point of BMBM and ACAC. KK is the centroid of △ABC\triangle ABC.
7. The centroid divides each median in the ratio 2:12:1. Therefore:
AKKC=2  ⟹  AK=2x23,KC=x23 \frac{AK}{KC} = 2 \implies AK = \frac{2x\sqrt{2}}{\sqrt{3}}, \quad KC = \frac{x\sqrt{2}}{\sqrt{3}}
8. Using the Law of Cosines in △KMC\triangle KMC:
KM2=KC2+CM2−2⋅KC⋅CM⋅cos⁡75∘ KM^2 = KC^2 + CM^2 - 2 \cdot KC \cdot CM \cdot \cos 75^\circ
We know cos⁡75∘=6−24\cos 75^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}, so:
KM2=(x23)2+x2−2⋅x23⋅x⋅6−24 KM^2 = \left(\frac{x\sqrt{2}}{\sqrt{3}}\right)^2 + x^2 - 2 \cdot \frac{x\sqrt{2}}{\sqrt{3}} \cdot x \cdot \frac{\sqrt{6} - \sqrt{2}}{4}
Simplifying the expression:
KM2=2x23+x2−x2(12−6)23 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{12} - \sqrt{6})}{2\sqrt{3}}
KM2=2x23+x2−x2(23−6)23 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(2\sqrt{3} - \sqrt{6})}{2\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
KM2=2x23+x2−x2(3−2)3 KM^2 = \frac{2x^2}{3} + x^2 - \frac{x^2(\sqrt{3} - \sqrt{2})}{\sqrt{3}}
\[
KM^2 =

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.